The space of characters: some properties and examples

Fa oltà di S ienze Matemati he Fisi he e Naturali

Corso di Laurea Spe ialisti a

in Matemati a

Anno A ademi o 2008/2009

TESI DI LAUREA

The Spa e of Chara ters:

Some Properties and Examples

Candidato

Ni ola Pi o o

Relatore Controrelatore

Introdu tion 3

1 Tools from Algebrai Geometry 5

1.1 Some Elementary Notions of Algebrai Geometry . . . 5

1.2 Zariski's MainTheorem and itsConsequen es . . . 8

1.3 Notions about Algebrai Plane Curves . . . 12

1.3.1 Plane Curves . . . 19

1.4 Resolutionof Singularities . . . 28

2 Varieties of Group Representations 39 2.1 The Spa e of Representations . . . 40

2.2 Curves of Representation . . . 43

2.3 The Spa e of Chara ters . . . 45

2.3.1 S hur's Lemmaand Wedderburn's Theorem . . . 47

2.3.2 The Closed Algebrai Set

χ(Π)

. . . 58

2.4 RelationAbout Dimensions . . . 62

2.5 Appendix . . . 65

3 Examples of Spa es of Chara ters 67 3.1 Presentation of a Group . . . 67

3.2 The Spa e of Chara ters of Some 1-DimensionalSpa es . . . . 69

3.2.1 The Spa e of Chara ters of

S

1

. . . 70

3.2.2 The Spa e of Chara ters of the Bouquet of Two Cir les 73 3.3 The Spa e of Chara ters of Some 2-DimensionalSpa es . . . . 74

3.3.1 The Spa e of Chara ters of the Proje tive RealPlane . 75 3.3.2 The Spa e of Chara ters of the Pun tured Torus . . . . 76

3.3.3 The Spa e of Chara ters of the Torus . . . 76

3.3.4 The Spa e of Chara ters of the Klein Bottle . . . 80

The purpose of this thesis isto give apresentation of the spa e of hara ters

of a nitely generated group

Π

. This is obtained by onsidering rst of all the representations of the group in

SL(2; C)

: this gives a spa e, alled the spa e of representations, that is easily shown to be a losed algebrai set in

C

4n

, where

n

is the number of generatorsof the group

Π

.For every element

g

_{∈ Π}

, one an onsider the fun tion on

R(Π)

whi h asso iates to every point(arepresentation) the tra eof the imageof gunderthe representation

orresponding to the point.

We will show that the ring

T

formed by all these fun tions is nitely-generated [see (2.3.1)℄: the image of the appli ation from

R(Π)

to

C

k

ob-tained by the fun tions that generate

T

(for a suitable

k

), form the "spa e of hara ters", denoted by

χ(Π)

.

Also in this ase the so-obtained spa e turns out to be a losed

alge-brai set [see (2.3.22)℄. However the proof is de idedly more ompli ated in

omparison to the ase of the spa e of the representations. In literature it

is possible tond elementary but extremely long and ompli ated proofs of

this result (see for example [7℄). In this thesis we follow the proof given by

Culler-Shalenin[4℄,inwhi hmanyresultsfromalgebrai geometryareused.

These results are re alled in the rst hapter where in parti ular we show

that all the rational maps from a smooth urve to a proje tive variety are

regular (this result is based on Zariski's Theorem [see (1.2.7)℄); inthe same

hapter weprovethat everyplanar urveadmitsadesingularization[see 1.4℄

. These two results will be the key to prove in hapter 2 that the hara ter

spa e is a tually a losed ane variety.

The se ond hapter is devoted to dening the spa e of representations

and the spa e of hara ters as well as to prove that the se ond is a losed

algebrai variety.Besides the already ited results omingfromalgebrai

ge-ometry, we will need the Burnside lemma [see (2.3.19)℄ and Wedderburn's

theorem [see (2.3.18)℄.Bothresultswillbeproved insubse tionsofthe same

In the third hapter we will provide expli it examples of spa es of

har-a ters of groups and of surfa es (i.e. of fundamental groups of surfa es) by

providingaset ofdening equations.Wewillrst treat the ase ofthe ir le

[see 3.2.1℄ and of the bouquet of two ir les[see 3.2.2℄. Then wewillanalyze

the spa es of hara ters of some 2-dimensional surfa es like the proje tive

plane [see3.3.1℄, thepun tured torus[see 3.3.2℄,the torus[see 3.3.3℄and the

Tools from Algebrai Geometry

Purpose ofthis hapteristogivesome algebrai geometrytoolsuseful inthe

next hapters. Our attention will on entrate ontwo main results: one says

that every rational map from a smooth urve to a proje tive variety is

reg-ular; the other that for any algebrai urve there exists asmooth proje tive

variety birationaltoit, hen e with the same fun tion eld (for simpli ity in

this work will be proved the assertiononly for the plane urves and we will

refer the reader tothe bibliography for aproof in the general ase).

1.1 Some Elementary Notions of Algebrai

Ge-ometry

We re all in this se tion some denitions and onstru tion that we will use

in this hapter. We start re allingthe denition of algebrai variety.

Denition 1.1.1 (AneVariety)A losedalgebrai set

V (β) =

{x ∈ C

n

_{|f(x) =}

0,

∀f ∈ β}

, where

β

isa prime idealin

C[X

1

, . . . , X

n

]

, is alled an ane va-riety.

Remark 1.1.2

β

primeidealimpliesthatavarietyisirredu ible;i.e. annot be write as union of two distin t losed sets.

Denition 1.1.3 (Rational and Regular Map) Let be

X = V (β)

an ane variety, then we denote by

C(X)

the set of rational fun tion of

X

, i.e. the eldof fra tionsof

C[X] = C[X

1

, . . . , X

n

]/β

. Arationalfun tion

ϕ

isregular at

x

∈ X

if it an be written in the form

ϕ = f /g

with

f, g

∈ C[X]

and

g(x)

_{6= 0}

. A rational map

ϕ : X

→ Y

(where

Y

⊂ C

m

set) is a

m

-tuple of rational fun tions

ϕ

1

, . . . , ϕ

m

∈ C(X)

su h that, for all points

x

∈ X

atwhi h allthe

ϕ

i

are regular,

ϕ(x) = (ϕ

1

(x), . . . , ϕ

m

(x))

∈ Y

; we say that

ϕ

is regular at su ha point

x

.

Denition 1.1.4 (Birationaland BiregularMap) A rational map

ϕ : X

→ Y

is birational if

ϕ

has an inverse rational map

ψ : Y

→ X

, that is,

ϕ(X)

is dense in

Y

and

ψ(Y )

in

X

, and

ψ

◦ ϕ =

Id

X

,

ϕ

◦ ψ =

Id

Y

(where dened). In this ase we say that

X

and

Y

are birational. Moreover if both

ϕ

and

ψ

are regular then we say that

ϕ

is a biregular map.

Denition 1.1.5 (Proje tive Variety) A losed algebrai set

V (β)

, where

β

is a homogeneous prime idealin

C[X

0

, . . . , X

n

]

, is alled a proje tive variety. Denition 1.1.6 (Corresponden e) Let

X

⊂ P

n

and

Y

⊂ P

m

be two

vari-eties.A orresponden e

Z

from

X

to

Y

isarelationgivenbya losedalgebrai subset

Z

⊂ X × Y

.

Denition 1.1.7 (Rational and Birational Map, Proje tive Case) A

orre-sponden e

Z

is said to be a rational map if

Z

is irredu ible and there is a Zariski open set

X

0

⊂ X

su h that ea h

x

∈ X

0

is related by

Z

to one and only one point of

Y

.

Z

is said to be birational map if

Z

⊂ X × Y

and

Z

−1

₌

_{{(y, x)|(x, y) ∈ Z} ⊂ Y × X}

are both rational maps.

Denition 1.1.8 (RegularMap,Proje tiveCase)Let

Z

⊂ X×Y

bearational map from

X

to

Y

, where

X

and

Y

are proje tive varieties of dimension

n

and

m

respe tively. Let

x

∈ X

and denote

Z [x] =

{y ∈ Y | (x, y) ∈ Z}.

Then

Z

is regular at

x

if: i)

Z [x] = (

a single point

y)

ii) if

(X

1

, . . . , X

n

)

, risp.

(Y

1

, . . . , Y

m

)

areane oordinate around

x

, resp.

y

,theninsomeZariski neighborhood

U

of

x

,

Z

isthegraphofthemap:

U

_{−→ Y}

given by

Y

i

=

a

i

(X

1

, . . . , X

n

)

b

i

(X

1

, . . . , X

n

)

where

a

i

and

b

i

are polynomials su h that

b

i

is nowhere zero on

U

.

X

reg

Z

=

{x ∈ X|Z

regular at

x

}.

It followsby denition that itisZariski-open.Finally wewilldenotefor any

ane variety

X = V (β)

⊂ C

n

, where

β

is aprime ideal and for any

x

∈ X

O

x,X

=



ring of rationlfun tion

f (X

1

, . . . , X

n

)

g (X

1

, . . . , X

n

)

,

where

g (x)

6= 0,

modulothose with

f

∈ β

!

Similarly,for any proje tive variety

X = V (β)

⊂ P

n

,where

β

isa homo-geneous primeideal, and for any

x

∈ X

we willdenote:

O

x,X

=



ring of rationl fun tion

f (X

0

, . . . , X

n

)

g (X

0

, . . . , X

n

)

, f, g,

homogeneous

of the samedegree,

g (x)

6= 0,

modulo the ideal of

f /g

′

_{s, f}

_{∈ β}

!

In both ases

O

x,X

isa lo al ring and

M

x,X

=

{h ∈ O

x,X

|h(x) = 0}

is itsmaximal ideal.

Denition 1.1.9 (Derivation) Let

X

be a variety and

x

∈ X

. A derivation

D : C[X]

_{→ C}

entred at

x

is a

C

-linear map su h that i)

D(f g) = f (x)

· D(g) + g(x) · D(f)

;

ii)

D(z) = 0

,

∀z ∈ C

.

Fora variety (aneor proje tive)we dene

T

x,X

=

{

ve tor spa e of derivations

D :

O

x,X

→ C

entred at

x

},

wherethederivation

D

extendsuniquelytothering

O

x,X

bytherule

D(f /g) =

(g(a)Df

− f(a)Dg)/g(a)

2

,

∀a ∈ X

.

Denition 1.1.10 (Dimension of a Variety) Let

X

be a variety. The dimen-sion of

X

is dened by

dim

X =

tr.d.

C

C(X) =

1

min

_x∈X

dim

T

x,X

.

1

Denition 1.1.11 (Smooth Point) Let

X

be a variety. A point

x

∈ X

is alled smooth if dim

T

x,X

=

dim

X

.

Proposition 1.1.12 Theset of smoothpoints of

X

is anon-empty Zariski-open subset of

X

. (See [2℄ Proposition(1.12), page 5)

Denition 1.1.13 (Dominating Map) A regular map

ϕ : X

→ Y

of ane varieties is saiddominating if

ϕ (X)

is Zariski-densein

Y

, i.e.

ϕ (X) = Y

Denition 1.1.14 (SmoothMap) Let

ϕ : X

r

_{→ Y}

s

be a dominatingregular

map. Let

x

∈ X

and

y = ϕ (x)

, than we say that

ϕ

is smooth at

x

if: i)

x

and

y

are smooth points of

X

and

Y

respe tively;

ii)

dϕ

maps

T

x,X

onto

T

y,Y

. 2

Sin e by assumption dim

T

x,X

= r

and dim

T

y,Y

= s

, this is equivalent to saying thatKer(

dϕ

) is

(r

− s)

-dimensional.

Proposition 1.1.15 Theset of smooth points of

ϕ

is a non-empty Zariski-open subset of

X

. (See [2℄ Proposition(3.6), page 42)

1.2 Zariski's Main Theoremand its Consequen es

In this se tion we willprovethe following proposition:

Proposition 1.2.1 i) Arationalmapfromasmooth urve

X

toavariety

Y

isregular;

ii) A birational map between smooth urves is biregular.

In orderto prove this fa t we willneed of Zariski's MainTheorem.

We start stating awell-know resultand anits orollary.

Theorem 1.2.2 Let

ϕ : X

r

_{→ Y}

s

be a dominating regular map of ane

varieties of dimension

r

and

s

respe tively. 3

For all

y

∈ Y

, all omponents of

ϕ

−1

_(y)

have dimension at least

r

− s

. Moreover the points for whi h all omponents of

ϕ

−1

_(y)

have dimension exa tly

r

− s

form an open. (See [2℄ Theorem(3.13) and Corollary(3.15), page 45-46)

2

Foranexpli itdenitionof

dϕ

,see [2℄, Denition(3.2),page41. 3

By

X

n

wewill denotealwaysa

n

-dimensionalvariety,andwewill state aseby ase whether

X

isaneorproje tive.

Corollary 1.2.3 If

X

is a

r

-dimensional ane variety and

f

1

, . . . , f

k

are polynomial fun tions on

X

, then every omponent of

X

∩ V (f

1

, . . . , f

k

)

has dimension

≥ r − k

. (See [2℄ Corollary(3.14), page 46)

We willneed of the followingdenition:

Denition 1.2.4 (Regular Corresponden e) Let

ϕ : X

r

_{→ Y}

s

be a regular

map. Let

X

1

, . . . , X

n

resp.

Y

1

, . . . , Y

m

be ane oordinates on

X

, resp.

Y

. Let

x

∈ X

and assume

y = ϕ(x)

. Then

ϕ

−1

is a regular orresponden e at

y

if

∃

a polynomial

d (Y

1

, . . . , Y

m

)

su h that

d(y)

6= 0

and an inverse

ψ : Y

′

=

_{{y ∈ Y |d(y) 6= 0} −→ X}

to

ϕ

dened by

X

i

=

a

i

(Y

1

, . . . , Y

m

)

d (Y

1

, . . . , Y

m

)

for some polynomials

a

i

in

C[Y

1

, . . . , Y

m

]

.

Theorem 1.2.5 Zariski's MainTheorem(AneSmoothCase).Let

X

r

_⊂

C

n

and

Y

r

_{⊂ C}

m

be two ane varieties of the same dimension

r

with

X

1

, . . . , X

n

and

Y

1

, . . . , Y

m

respe tively as ane oordinates, and let

ϕ :

X

_{→ Y}

a birational regular map between them. Let

x

∈ X

and assume

y = ϕ (x)

is smooth on Y. Then either a)

ϕ

−1

is a regular orresponden e at

y

; or

b)

∃

asubvariety

E

⊂ X

through

x

ofdimension

r

−1

su hthatdim

ϕ (E)

≤

r

− 2

.

In parti ular,

ϕ

−1

_(y)

has a positive-dimensional omponentthrough

x

. Proof It is well-knowthat if

ϕ

is birational,the pull-ba k

ϕ

∗

,dened by

ϕ

∗

_{: K (Y )}

_{−−−→ K (X)}

f

_{−−−→ f ◦ ϕ}

isanisomorphism,thenea hfun tion

X

i

on

X

equals

a

i

(Y )

b

i

(Y )

◦ϕ

forsome

poly-nomials

a

i

, b

i

(

b

i

6= 0

on

Y

). Another well-know fa t is that if

y

is smooth on

Y

then

O

y,Y

is a Unique Fa torization Domain (UDF) (see [2℄ Proposi-tion(1.16), page 15). It follows that we an write:

X

i

=

a

i

(Y )

b

i

(Y )

◦ ϕ

Nowtwo ases are possible:

a) for all

i

,

b

i

(y)

6= 0

: in this ase it is enoughto put

d =

Q

b

i

and dene

ψ

by

X

i

=

a

i

(Y )

·

Q

j6=i

b

j

(Y )

d (Y )

;

b) at least one

b

i

vanishes at

y

: say

b

1

(y) = 0

. Let

β (Y

1

, . . . , Y

m

)

be a polynomialthat represents an irredu ible fa tor of

b

1

in

O

y,Y

; say

b

1

= b

′

1

β

. Take for

E

a omponent of

X

∩ V (β ◦ ϕ)

through

x

. By Corollary(1.2.3), dim

E = r

− 1

. But on

X

we have

a

1

◦ ϕ = X

1

· b

′

1

◦ ϕ

· (β ◦ ϕ) ,

hen e

a

1

◦ ϕ = 0

on

E

.Therefore

a

1

= β = 0

on

ϕ (E)

.

Sin e

β

isirredu ible

β

· O

y,Y

is aprimeideal,and then

B =

{f ∈ C [Y ] |f ∈

β

· O

y,Y

}

is a primeideal also in

C

[Y ]

. Moreover

a

1

∈ B

/

sin e

a

1

and

b

1

are relatively prime,hen e

a

1

6≡ 0

on

V (B)

. So wend

Y ) V (B) ) ϕ (E)

hen e dim

ϕ (E)

≤ r − 2

.



We nowgoba kto proje tive varietiesand orresponden es

Z

⊂ X × Y

. Westart toobservethat the proje tions

p

1

: Z

−→ X

p

2

: Z

−→ Y

are maps that are lo ally proje tions

C

n+m

_{→ C}

n

. In parti ular the above

ane resultapply toit.Thus, suppose

p

1

(Z) = X

,

p

2

(Z) = Y

and onsider the fun tions:

f (x, y) =

max

{

dim

W

|W

a omponentof

Z [x]

through

(x, y)

}

and

f (x) =

max

{

dim

W

|W

a omponentof

Z [x]

}.

Sin e

p

1

: Z

→ X

isgeneri allysmooth,the valueof

f (x)

almosteverywhere is dim

Z

−

dim

X

.Therefore we may de ompose

X

into two pie es:

1) anon-empty Zariski-open pie e

X

0

whereall omponentsof

Z [x]

have dimensiondim

Z

−

dim

X

(thefa tthatthisisindeedanopensetfollows by Theorem(1.2.2));

2) aZariski- losedpie e

F

(the omplementof

X

0

)wheresome omponent of

Z [x]

has largerdimension:theseare alledthe fundamentalpointsof

Z

.

Remark 1.2.6 The odimension in

X

of all omponentsof

F

is at least 2. Proof Lookat:

Z

_{⊃ F}

∗

₌

_{{(x, y) |f (x, y) >}

dim

Z

−

dim

X

}

p

_{1|F ∗}

−−−−−−−−→ F

then all omponents of

F

∗

havedimension atmost dim

Z

− 1

whileall bers of res

p

1

havedimensionatleast dim

Z

−

dim

X + 1

,thusall omponentsof

F

have dimension atmost dim

X

− 2

.



This tells us that

X

reg

is a non-empty Zariski-open subset of

X

disjoint from

F

.Then the proje tive formof Zariski's Main Theorem follows:

Theorem 1.2.7 (Zariski's Main Theorem) (Proje tive-SmoothCase)Let

X

and

Y

be proje tive varieties and let

Z

⊂ X × Y

be a rational map from

X

onto

Y

. Then

X

− F −

Sing

X

⊂ X

Z

reg

,

i.e.,

Z

is regular at every smooth non-fundamental point.

Proof First of all we observe that sin e

Z

is a rational surje tive map, dim

X

=dim

Z

.Indeed, suppose dim

X = s

, and let

X

1

, . . . , X

s

bea trans en-den e base of

C(X)

over

C

(where the

X

i

are the oordinate fun tions on

X

);letdim

Y = r

,and let

Y

1

, . . . , Y

r

beatrans enden e baseof

C(Y )

over

C

(where the

Y

i

are the oordinatefun tions on

Y

). It is enough toshow that if we add to

X

1

, . . . , X

s

one of

Y

i

, these are algebrai allydependent over

Z

. Suppose

Z =

{(x, ϕ(x))|x ∈ X}

, and suppose that the

i

-th omponent of

ϕ

is inthe form

ϕ

i

=

a

b

.Then the polynomial

q(X

1

, . . . , X

s

, Y

i

) = b(X

1

, . . . , X

s

)

· Y

i

− a(X

1

, . . . , X

s

)

is not identi allyzero but, by surje tivity, vanishes overall pointsof

Z

. From this in parti ular it follows that the subset

F

⊂ X

of fundamental pointsis su h that

∀x ∈ F

dim

Z[x]

≥ 1

; in parti ular we nd again that in this points

Z

annotberegular sin eit doesnot satised the ondition i)of denition of regular map.

Now onsider the proje tionmap

p

1

: Z

→ X

restri ted toanane open set

U

su h that

p

1

(U)

⊂ X − F −

Sing

X

: we are in the hypothesis of the

ane versiontheorem, so wehave that forany

x

∈ X − F −

Sing

X

itexists a regularinverse of

p

1

,say

ψ : X

→ Z

,in the formdened by

X

i

= X

i

∀i = 1, . . . , n

and

Y

j

=

a

j

(X

1

, . . . , X

n

)

d(X

1

, . . . , X

n

)

∀j = 1, . . . , m,

where the

a

j

and

d

are polynomials on

X

, and

ψ

is dened over the points in whi h

d

does not vanish. Thus for any

x

∈ X − F −

Sing

X

we see that map

Z

isthe graphi , ona neighborhood

U

of

x

,of the map

U

−→ Y

given by

Y

i

=

a

i

(X

1

, . . . , X

n

)

d (X

1

, . . . , X

n

)

,

i.e.

Z

is aregularmap ateverysmooth non-fundamentalpointsasrequired.



We are now ready for to state the purpose of this se tion that is an

im-mediate onsequen e of Zariski's MainTheorem in the proje tive ase.

Proof (of 1.2.1) Obviously (ii)follows from (i).

As for (i):if we denote the urve by

X

and the map by

Z

, then Sing

X =

∅

sin e the urve issmooth and

F =

∅

sin e dim

X = 1

and by Remark(1.2.6). Therefore

X = X

Z

reg

.



1.3 Notions about Algebrai Plane Curves

Now we fa e to the se ond goal of this hapter, i.e. that for any algebrai

urve there exists a smooth proje tive variety birationaltoit.

Start with asimple denition.

Denition 1.3.1 (Fun tion Field) A eld

K

ontaining

C

and nitely gen-erated over it is alled a fun tion eld.

Remark 1.3.2 For ea h fun tion eld

K

there exists a variety

X

su h that

Proof Let

K

be a eld of trans enden e degree

n

, and

x

1

, . . . , x

n

∈ K

a trans enden e base.Then

K

isanitealgebrai extensionof

C(X

1

, . . . , X

n

)

, and by primitive element theorem su h extension an be generated by one

element; say

x

n+1

. Then in

K x

1

, . . . , x

n

, x

n+1

must satisfy an irredu ible equation

f (X

1

, . . . , X

n+1

) = 0

, unique up to s alars.Therefore

K =

fra tion eld of

C[X

1

, . . . , X

n+1

]/(f ) = C(X)

where

X

⊂ P

n+1

is the hypersurfa e given by

X = V (f )

.



Denition 1.3.3 (Model of Fun tion Field) When

K = C(X)

and

X

is a proje tive variety, we say that

X

is a model of

K

.

Itis learthat if

X

1

and

X

2

aretwomodelsof thesame eld

K

,then the omposition

C(X

1

)

←−−− K

≈

−−−→ C(X

≈

2

)

denes a birational orresponden e between

X

1

and

X

2

.

Anaturalquestionatthis pointis:given

K

,doesithaveasmoothmodel

X

?

If

K

is the fun tion eld of a urve, then the answer is positive. Indeed it holds the following result:

Theorem 1.3.4 For every

K

of trans enden e degree

1

, a smooth model exists.

4

Reallywewillprovethe theoremonlyforplane urves. Foraproofinthe

general ase see [2℄ Theorem(7.5), page 129.

In this ase the statementof the theorem be omes:

Theorem 1.3.5 Let

K

the fun tion eld of a plane urve in

P

2

; then it

exists a smooth urve in

P

N

, for a suitable

N

, with the same fun tion eld

K

.

In this se tion we will give the toolsthat we willuse in the next se tion

for the proof of Theorem(1.3.5).

4

Inparti ularweknowthatthis impliesthatavarietyhaving

K

asfun tion eld has dimensionequalto

1

.

C

n

is a polynomial map

T = (T

1

, . . . , T

n

) : C

n

_{→ C}

n

su h that

T

is one-to-one and onto.

Wewilluse thefollowingnotations:given anane hangeof oordinates

T

, if

F

∈ C

n

_[X

1

, . . . , X

n

]

is a polynomial then

F

T

_{= ˜}

_{T (F ) = F (T}

1

, . . . , T

n

)

; if

β

is an ideal in

C

n

_[X

1

, . . . , X

n

]

and

X = V (β)

⊂ C

n

an algebrai set, then

β

T

denotetheideal generatedby

{F

T

_{|F ∈ β}}

and

X

T

thealgebrai set

T

−1

_{(X) = V (β}

T

₎

.

Remark 1.3.7 Imposinglinear onditionsitiseasytoseethat,if

x, x

′

_{∈ C}

n

,

l

1

, l

2

are two distin tlines trough

x

and

l

′

1

, l

2

′

are two distin t lines trough

x

′

then there exist an ane hange of oordinates

T

su h that

T (x) = x

′

and

T (l

i

) = l

′

i

for

i = 1, 2

.

Lemma 1.3.8 Let

T : C

n

_{→ C}

n

be an ane hange of oordinates and

suppose

T (x) = y

, where

x

and

y

are two points of

C

n

. Then

T :

˜

O

y,C

n

→

O

x,C

n

is an isomorphism. Moreover, if

X

⊂ C

n

an algebrai set and

x

∈ X

then

T :

˜

O

y,X

T

→ O

_x,X

isan isomorphism too.

Proof Clearly

T (λf + f g) = λ ˜

˜

T (f ) + ˜

T (f g) = λ ˜

T (f ) + ˜

T (f ) ˜

T (g)

,forany

λ

∈ C

and

f, g

∈ O

y,C

n

.

If

f = l/h

with

h(y)

6= 0

, then

T (h)(x) = h(T )(x) = h(y)

˜

6= 0

; so

T (f )

˜

∈

O

x,C

n

. Similarly for

g

.

Inje tivity: if

T (F ) = ˜

˜

T (G)

,then

F

◦ T = G ◦ T

, so

F = G

;

Surje tivity:let

G

∈ O

x,C

n

,andsin e

T

isinvertiblewe antake

F = G

◦T

−1

.

Finally,if

x

∈ V

,the same argument works for

T :

˜

O

y,X

T

→ O

_x,X

.



Remark 1.3.9 Let

L

1

, L

2

, . . .

and

M

1

, M

2

, . . .

be sequen esof linear forms 5

in

C[X, Y ]

, and assume no

L

i

= γM

j

,

γ

∈ C

(note that possibly ea h se-quen es an be formed by forms all equal ones among them). Let

A

ij

=

Q

i

m=1

L

m

·

Q

j

n=1

M

n

,

i, j

≥ 0

(

A

00

= 1

). Then

{A

ij

|i + j = d}

forms a basis for

{

form of degree

d

in

C[X, Y ]

}

.

Proof It is easy to see that

{X

i

_Y

j

_{|i, j ≥ 0, i + j = d}}

is a base of

{

form of degree

d

in

C[X, Y ]

}

, so it is enough to show that any monomial

X

i

Y

j

an be writtenas a linear ombinationof

A

ij

.

By indu tion. Sin e we have put

A

00

= 1

, the assertion for

d = 0

is obvious. 5

For

d = 1

we have

X = α

i

L

i

+ β

j

M

j

(with

α

i

, β

j

∈ C

) for all

i

∈ N

sin e

L

i

6= γM

j

for any

γ

∈ C

. Suppose now the statement true for

d

− 1

; let

X

i

_Y

j

be a monomial of degree

d

− 1

(

i + j = d

− 1

) and we an suppose

i

6= 0

,sowe anwrite

X

i

_Y

j

_{= X(X}

i−1

_Y

j

₎

.Byindu tionhypothesiswe have

X

i−1

_Y

j

₌

P

i+j=d−1

a

ij

A

ij

.Hen e

X

i

Y

j

= X(X

i−1

Y

j

) =

X

i+j=d−1

a

ij

A

ij

X =

X

i+j=d−1

a

ij

A

ij

(α

i+1

L

i+1

+β

j+1

M

j+1

)

as desired.



Denition 1.3.10 (OrderFun tion,orDis reteValuation)AnOrderFun tion

(or Dis rete Valuation) on a eld

K

is a fun tion

ϕ

from

K

onto

Z

∪ ∞

, satisfying:

i)

ϕ(a) =

∞

i

a = 0

; ii)

ϕ(ab) = ϕ(a) + ϕ(b)

;

iii)

ϕ(a + b)

≥ min(ϕ(a), ϕ(b))

.

Note that the subset

R =

{z ∈ K|ϕ(z) ≥ 0}

is a Noetherian and lo al ring with maximal ideal

M =

{z ∈ K|ϕ(z) > 0}

, and quotient eld

K

.

Denition 1.3.11 (Dis rete Valuation Ring) Let

ϕ

an order fun tion. The ring

R =

{z ∈ K|ϕ(z) ≥ 0}

is all a dis rete valuation ring (DVR).

Proposition 1.3.12 If

R

is a lo al Noetherian domain and the maximal ideal is prin ipal, then there exist an irredu ible element

t

∈ R

su h that every non-zero

z

∈ R

may be written uniquely in the form

z = ut

n

, with

u

a unit in

R

and

n

a non-negative integer.

Proof Let

M

be the maximalideal and

t

anits generator. Uniqueness: let

ut

n

_{= vt}

m

, where

u, v

units and suppose

n

≥ m

; then

ut

n−m

_{= v}

is aunit, so

n = m

and

u = v

.

Existen e let

z

∈ R

not a unit, so

z = z

1

t

for some

z

1

∈ R

. Now either

z

1

is a unit or

z

1

= z

2

t

for some

z

2

∈ R

. In this way we get an innite sequen e

z

1

, z

2

, . . .

with

z

i

= z

i+1

t

. Sin e

R

isNoetherian,the hain ofideals

(z

1

)

⊂ (z

2

)

⊂ . . .

must havea maximal element;so

(z

n

) = (z

n+1

)

.



An element

t

as in the proposition is alled a uniformizing parameter for

that is prin ipal is a dis rete valuation ring.

Proof If we dene ord

(0) =

∞

, we have that ord:

K

→ Z ∪ {∞}

is an order fun tionon

K

,where

K

is the quotient eldof

R

.Indeed, if

t

is xed, it is immediatelyto verify that the fun tion "ord" satisesthe properties of

the order fun tion. Moreover it is independent of the hoi e of uniformizing

parameter:if

t

1

and

t

2

are twodistin tuniformizingparameters, thenonthe one hand

t

2

= u

2

t

s

1

, with

s

anon-negativeinteger and

u

2

aunit in

R

; but on the other hand

t

1

= u

1

t

r

2

, with

r

a non-negative integer and

u

1

a unit in

R

. Thus

t

2

= u

2

u

s

1

t

rs

1

, so

rs = 1

and then

r = s = 1

. This implies that for all

z

∈ R

we have

z = ut

n

1

= vt

m

2

, with

u, v

unit in

R

, but

t

2

= wt

1

(

w

unit in

R

) so

vt

n

2

= vw

n

t

n

1

and then

n = m

by uniqueness.



We provided now some lemmas about ideals and relationsbetween

alge-brai sets and ring homomorphisms.

Lemma 1.3.14 (i) Let

I

⊂ J

beidealsinaring

R

.Thenthereisanatural homomorphism from

R/I

onto

R/J

.

(ii) Let

I

be an idealin a ring

R

, with

R

a subring of a ring

S

. Then there is a natural homomorphism from

R/I

to

S/IS

.

Proof (i)Let

d

∈ R/I

and

d+i

with

i

∈ I

anitsrepresentativein

R

.Sin e

I

_{⊂ J}

we an write

d = d

′

_{+ j}

. Then we set

ϕ : R/I

→ R/J

by

ϕ(d) = d

′

. This is learly an homomorphismand it is"onto" sin e for all

d

′

∈ R/J

we have that

d

′

is alsoanelement of

R/I

.

(ii) Let

a

∈ R/I

and

a + i

with

i

∈ I

an its representer in

R

. But

a + i

is alsoanelement of

S

,so we an write

a + i = b + js

,with

j

∈ I

,

s

∈ S

.Then we set

ϕ : R/I

→ S/IS

by

ϕ(a) = b

. It is immediate to see that this is an homomorphism.



Lemma 1.3.15 Let

x = (0, . . . , 0)

∈ C

n

,

O = O

x,C

n

,

M =

M

x,C

n

and

I = (X

1

, . . . , X

n

)

⊂ C[X

1

, . . . , X

n

]

. Then

I

O = M

, so

I

r

_O

_{= M}

r

for all integer

r

. Proof

M =

{f ∈ O

p

(C

n

₎

_{|f(p) = 0}}

, while

I

O = {

P

x

i

h

i

with

h

i

∈ O}

. Weprove the double in lusion.

I

O ⊂ M

: learly

P

x

i

h

i

∈ O

and itvanish at

p

;

h(p) = 0

;then

h

hasnot onstantterms,sowe anwriteitas

P

i∈{1,...,n}

x

i

s

i

; thus

f = h/l =

P

i∈{1,...,n}

x

i

(s

i

/l)

∈ IO

.



Lemma 1.3.16 Let

X

beavarietyin

C

n

,

I = I(X)

⊂ C[X

1

, . . . , X

n

]

,x

∈ X

, and let

J

be an ideal of

C[X

1

, . . . , X

n

]

whi h ontains

I

. Let

J

′

be the image

of

J

in

C[X]

. Then there is a natural isomorphism

ϕ :

O

x,C

n

J

O

x,C

n

−→

O

x,X

J

′

_O

_x,X

.

In parti ular,

O

x,Cn

IO

x,Cn

isisomorphi to

O

x,X

. Proof If

[l]

is anequivalent lass in

O

x,Cn

JO

x,Cn

, then a representative of

[l]

in

O

x,C

n

has the form

l + jm

, with

j

∈ J

and

m

∈ O

x,C

n

.Let

π : C[X

1

, . . . , X

n

]

−→

C[X

1

, . . . , X

n

]

I

bethenaturalhomomorphism,andlet

l +jm

betheimageof

l +jm

under

π

, wherebarsdenotes the

I

-residue.Sin e

j

∈ J

′

,then

jm

∈ J

′

_O

x,X

.Therefore it is enoughtoset

ϕ([l]) = [l]

.



Remark 1.3.17 Two ideals

I, J

⊂ C[X

1

, . . . , X

n

]

are omaximal (i.e.

I +

J = (1)

) if and only if

V (I)

∩ V (J) = ∅

.

Proof If

I + J

6= (1)

then

V (I)

∩ V (J) = V (I + J) 6= V (1) = ∅

: on-tradi tion; onversely, if

∅ 6= V (I) ∩ V (J) = V (I + J)

, then

∃ p

su h that

∀ f ∈ I, ∀ g ∈ J

we have

f (p) + g(p) = 0

, thus

∀ f ∈ I, ∀ g ∈ J

it holds

f + g

6= 1

: ontradi tion.



Remark 1.3.18 Let

I, J

beidealsinaring

R

.Suppose

I

isnitelygenerated and

I

⊂

Rad

(J)

, then

I

n

_{⊂ J}

for some

n

.

Proof Suppose

I

is generated by

{x

1

, . . . , x

m

}

; then

∀ i ∃ n

i

su h that

x

n

i

i

∈ J

.I laimthat

n =

P

n

i

is thedesire number. Indeed,inea helement

g

of

I

n

they appear the terms

P

ax

s

1

1

. . . x

s

m

m

with

a

∈ R

and

P

s

i

= n

, so at least one of these

s

i

must be greaterthan

n

i

,thus

g

is in

J

.



Remark 1.3.19 Let

{P

1

, . . . , P

r

}

be a nite set of points in

C

n

. Then there

are polynomials

F

1

, . . . , F

r

∈ C[X

1

, . . . , X

n

]

su hthat

F

i

(P

j

) = 0

if

i

6= j

and

Proof Set

V

i

=

∪

j6=i

P

j

. Sin e

V

i

V

i

∪ {P

i

}

, then

I(V

i

∪ {P

i

}) I(V

i

)

; thus it exists

G

i

∈ I(V

i

)

but

G

i

6∈ I(V

i

∪ {P

i

})

. It follows that

G

i

(P

i

)

6= 0

, thus

F

i

= G

i

/G

i

(P

i

)

isthe desired polynomial.



The next propositionis an useful tool.

Proposition 1.3.20 Let

I

beanidealin

C[X

1

, . . . , X

n

]

,andsuppose

V (I) =

{x

1

, . . . , x

N

}

is nite. Let

O

i

=

O

x

i

,C

n

. Then there is a natural isomorphism

ϕ :

C[X

1

, . . . , X

n

]

I

−→ ×

N

i=1

O

i

I

O

i

.

Proof Let

I

i

= I(

{P

i

}) ⊂ C[X

1

, . . . , X

n

]

be the distin t maximal ide-als whi h ontain

I

, and let

R = C[X

1

, . . . , X

n

]/I

,

R

i

=

O

i

/I

O

i

. By the Lemma(1.3.14)(b), it follows that the natural homomorphism

ϕ

i

from

R

to

R

i

indu ea homomorphism

ϕ

from

R

to

×

N

i=1

R

i

.

Bythe Nullstellensatzand Remark(1.3.18),Rad

(I) = I(

{P

1

, . . . , P

N

}) =

T

N

i=1

I

i

, so

(

T

I

i

)

d

⊂ I

for some

d

. By Remark(1.3.17),

T

j6=i

I

j

and

I

i

are omaximal. So, sin e powers of omaximal ideals are still omaximal, and

for omaximal ideals itholds that

I

∩ J = I · J

(and indu tivelyfor a nite numberof theme), itfollows that

T

(I

d

j

) = (I

1

· · · I

N

)

d

= (

T

I

j

)

d

⊂ I

. Now as in the Remark(1.3.19) hoose

F

i

∈ C[X

1

, . . . , X

n

]

su h that

F

i

(P

j

) = 0

if

i

6= j

,

F

i

(P

i

) = 1

. Let

E

i

= 1

− (1 − F

i

d

)

d

.

Note that

E

i

= F

d

i

D

i

for some

D

i

, so

E

i

∈ I

d

j

if

i

6= j

, and

1

−

P

i

E

i

=

(1

− E

j

)

−

P

_i6=j

E

i

∈

T

I

j

d

⊂ I

. Let

e

i

be the residue of

E

i

in

R

; then we have

e

2

i

= e

i

,

e

i

e

j

= 0

if

i

6= j

,and

P

e

i

= 1

.

Suppose

G

∈ C[X

1

, . . . , X

n

]

, and

G(P

i

)

6= 0

(sowe an assume

G(P

i

) =

1

), and let

g

be itsresidue in

R

. Let

H = 1

− G

. Wehave:

(1

− H)(E

i

+ HE

i

+

· · · + H

d−1

E

i

) = E

i

− H

d

E

i

.

Sin e

H

∈ I

i

,itfollowsthat

H

d

_E

i

∈ I

.Therefore

g(e

i

+he

i

+

· · ·+h

d−1

_e

i

) = e

i

. Thuswehaveprovedthatinthishypothesisitexists

t

∈ R

su hthat

tg = e

i

.

Usingthis result we nowprove that

ϕ

is anisomorphism.

Inje tivity: if

ϕ(f ) = 0

, where

f

is the residue of

F

in

R

, then for ea h

i

there is a

G

i

with

G

i

(P

i

)

6= 0

and

G

i

F

∈ I

. Let

g

i

be the residueof

G

i

in

R

, and take

t

i

asabove su h that

t

i

g

i

= e

i

. Then

f =

X

e

i

f =

X

Surje tivity:sin e

E

i

(P

i

) = 1

,itfollows that

ϕ

i

(e

i

)

isaunit in

R

i

,and sin e

ϕ

i

(e

i

)ϕ

i

(e

j

) = ϕ

i

(e

i

e

j

) = 0

if

i

6= j

, it follows that

ϕ

i

(e

j

) = 0

if

i

6= j

. Therefore

ϕ

i

(e

i

) = ϕ

i

(

P

e

j

) = ϕ

i

(1) = 1

.Let

z =



a

1

s

1

, . . . ,

a

N

s

N



∈ ×

N

i=1

R

i

.

Byaboveresultwemayset

t

i

s

i

= e

i

;then

a

i

/s

i

= a

i

t

i

in

R

i

.So

ϕ

i

(

P

t

j

a

j

e

j

) =

ϕ

i

(t

i

a

i

) = a

i

/s

i

, and

ϕ(

P

t

j

a

j

e

j

) = z

.



Corollary 1.3.21 Let

I

be an ideal in

C[X

1

, . . . , X

n

]

, and suppose

V (I) =

{x

1

, . . . , x

N

}

is nite. Let

O

i

=

O

x

i

,C

n

. Then dim

C



C[X

1

, . . . , X

n

]

I



=

N

X

i=1

dim

C



O

i

I

_O

i



.

Corollary 1.3.22 Let

I

be an ideal in

C[X

1

, . . . , X

n

]

. If

V (I) =

{x}

, then

C[X

1

, . . . , X

n

]/I

is isomorphi to

O

x,C

n

/I

O

x,C

n

.

1.3.1 Plane Curves

Afterre allingthedenitionofthenotionofaneplane urve,wewilldene

some on epts whi h give informationabout their geometri al proprieties.

Wesay thattwopolynomials

F, G

∈ C[X, Y ]

areequivalentif

F = λG

for some non-zero

λ

∈ C

Denition 1.3.23 (AnePlaneCurve)We dene anane plane urveto be

anequivalent lassofnon- onstantpolynomialsunderthisequivalentrelation.

The degree of a urve is the degree of a dening polynomial for the urve.

If

F

∈ C[X, Y ]

is irredu ible then

V (F )

is a variety in

C

2

. When this

don't will reate onfusion, we willdenote by

F

both the equation and the variety

V (F )

.

Let

F

bea urve,

P = (a, b)

∈ F

.

P

is alledasimple point of

F

if either derivative

F

X

(P )

6= 0

or

F

Y

(P )

6= 0

. In this ase the line

F

X

(P )(X

− a) +

F

Y

(P )(Y

− b) = 0

isthe tangent lineto

F

at

P

.A point whi h isnot simple is alled singular. A urve is said non-singular if all itspoint are simple.

Now let

F

be a urve and

P = (0, 0)

. Write

F = F

m

+ F

m+1

+

· · · + F

n

, where

F

i

is aformin

C[X, Y ]

of degree

i

,

F

m

6= 0

.We all

m

the multipli ity of

F

at

P

, and we write

m = m

P

(F )

.

Remark 1.3.24 Note that

m

P

(F ) = 0

if andonlyif

P /

∈ F

and

m

P

(F ) = 1

if and only if

P

isa simplepoint on

F

.

Write

F

m

=

Q

L

r

i

i

wherethe

L

i

are distin t lines, i.e.fa tors of the type

(αX

− βY )

. Then

L

i

are alled the tangent lines to

F

at

P

, and

r

i

is the multipli ityof thetangent;

L

i

isasimpletangentif

r

i

= 1

.If

F

has

m

distin t (simple) tangents at

P

,wesay that

P

is anordinarymultiple point of

F

.

If

P = (a, b)

6= (0, 0)

then we onsider the hange of oordinates given by

T (x, y) = (x + a, y + b)

, and we dene

m

P

(F ) = m

(0,0)

(F

T

₎

. If

F

T

₌

G

m

+ G

m+1

+

· · · + G

n

,

G

i

forms, then

m = m

P

(F )

. If

G

m

=

Q

L

r

i

i

,

L

i

= α

i

X + β

i

Y

, then the lines

α

i

(X

− a) + β

i

(Y

− b) = 0

are dened to be the tangent lines to

F

at

P

. At this point we an dene all the other on epts dened for the ase

P = (0, 0)

.

It follows a hara terization of simple point

P

on a urve

F

interm of a lo alring

O

P,F

.

Wewilluse the followingnotation: if

G

∈ C[X, Y ]

isany polynomial,then

g

denotes itsresidue image in

C[F ]

.

Remark 1.3.25 Let

I = (X, Y )

⊂ C[X, Y ]

. Sin e a base for

C

[X,Y ]

I

n

is given by

{x

i

_y

j

_{|i + j = 0, . . . , n − 1}}

then dim

C



C[X, Y ]

I

n



= 1 + 2 +

· · · + n =

n(n + 1)

2

.

Lemma 1.3.26 (1) Let

0

_{−−−→ V}

′

_{−−−→ V}

ψ

_{−−−→ V}

ϕ

′′

_{−−−→ 0}

be an exa t sequen eof nite-dimensional ve tor spa esover aeld

K

. Then dim

V

′

₊

dim

V

′′

₌

dim

V

. (2) Let

0

_{−−−→ V}

1

ϕ

1

−−−→ V

2

ϕ

2

−−−→ V

3

ϕ

3

−−−→ V

4

−−−→ 0

be an exa t sequen eof nite-dimensional ve tor spa esover aeld

K

. Then dim

V

4

=

dim

V

3

−

dim

V

2

+

dim

V

1

.

by (1)letting

W =

Im

(ϕ

2

) =

Ker

(ϕ

3

)

,and onsidering

0

_{−−−→ V}

1

ϕ

1

−−−→ V

2

ϕ

2

−−−→ W −−−→ 0

and

0

_{−−−→ W}

_{−−−→ V}

i

3

ϕ

3

−−−→ V

4

−−−→ 0

where

i

is the in lusion,and the result follows by subtra tion.



Remark 1.3.27 Let

R

be a dis rete valuation ring with maximal ideal

M

, and quotient eld

K

, and suppose a eld

k

is a subring of

R

, and that the omposition

k

−→ R −→ R/M

isan isomorphism of

k

with

R/M

. Itfollows that the

R

-module

M

n

_/M

n+1

is also a

k

-module. Moreover ea h element of

M

n

_/M

n+1

is in the form

[ax

n

_]

, where

a

∈ k

and

x

is a generator of

M

. In parti ular dim

k

(M

n

_/M

n+1

_{) = 1}

.

Remark 1.3.28 If

O

is a lo al ring with maximal ideal

M

, there exist a natural exa t sequen e of

O

-modules

0

−−−→

M

n

M

n+1

i

−−−→

O

M

n+1

π

−−−→

O

M

n

−−−→ 0,

sin e

O/M

n

isisomorphi to

(

O/M

n+1

_)/M

n

,thusKer

(π) = M

n

_/M

n+1

₌

Im

(i)

. Theorem 1.3.29 Let

P

be a point on a irredu ible urve

F

. Then

m

P

(F ) =

dim

C



M

p

(F )

n

M

P

(F )

n+1



for all su iently large

n

. In parti ular, the multipli ity of

F

at

P

depends only on the lo al ring

O

P,F

.

Proof Write

M,

O

for

M

P

(F ),

O

P,F

. ByRemark(1.3.28) the sequen e

0

_−−−→

_M

M

n+1

n

−−−→

_M

O

n+1

−−−→

_M

O

n

−−−→ 0

is exa t, and by Lemma(1.3.26)

dim

C

(

O/M

n+1

₎

−

dim

C

(

O/M

n

_{) =}

dim

C

M

p

(F )

n

_/M

P

(F )

n+1

,

soitisenoughtoprovethat dim

C

(

O/M

n

_{) = n}

_{· m}

p

(F ) + s

forsome onstant

We may assume that

P = (0, 0)

, so by Lemma(1.3.15)

M

n

_{= I}

n

_O

, where

I = (X, Y )

⊂ C[X, Y ]

. Sin e

V (I

n

_{) =}

_{{P }}

, by Proposition(1.3.20),

Corollary(1.3.22)and Lemma(1.3.16),itfollows that

C[X, Y ]

(I

n

_{, F )}

∼

=

O

P,C

2

(I

n

_{, F )}

_O

P,C

2

∼

=

O

P,F

I

n

_O

P,F

=

O

M

n

.

Let

m = m

P

(F )

. Then

F G

∈ I

n

whenever

G

∈ I

n−m

. If we dene

ψ :

C[X, Y ]/I

n−m

_{→ C[X, Y ]/I}

n

by

ψ(G) = F G

,where

G

is the residueof

G

,it follows that the sequen e

0

_−−−→

C

_I

[X,Y ]

n−m

ψ

−−−→

C

[X,Y ]

I

n

ϕ

−−−→

C

(I

[X,Y ]

n

_{,F )}

−−−→ 0

isexa t(where

ϕ

isthenaturalringhomomorphism).FinallybyRemark(1.3.25) and Lemma(1.3.26) again, we have dim

C

(C[X, Y ]/(I

n

_{, F )) = nm}

₋

m(m−1)

2

for all

n

≥ m

, as desired.



Theorem 1.3.30 Let

F

be a urve and let

P

∈ F

. Then

P

is a simple point of

F

if and only if

O

P,F

is a dis rete valuation ring. In this ase, if

L = aX + bY + c

is any line through

P

whi h isnot tangentto

F

at

P

, then the image

l

of

L

in

O

P,F

is a uniformizing parameter for

O

P,F

.

Proof ByRemark(1.3.7)andLemma(1.3.8),aftera hangeof oordinates,

we may assumethat

P = (0, 0)

,that

Y

is the tangentline, and that

L = X

; so by Proposition(1.3.12) and Corollary(1.3.13) it is enough to show that

x

generates

M

P

(F )

. First of all, note that by Lemma(1.3.15)

M

P

(C

2

_{) =}

(X, Y )

O

P,C

2

, and sin e

P = (0, 0)

∈ F

and hen e

(F )

⊂ (X, Y )

, then by Lemma(1.3.16) wehave:

C ∼

=

O

P,C

2

M

P

(C

2

)

∼

=

O

P,C

2

(X, Y )

_O

P,C

2

∼

=

O

P,F

(x, y)

_O

P,F

,

thus

M

P

(F ) = (x, y)

.

Now, by assumptions,we anwrite

F = Y +

higherterms,orequivalently

F = Y G

− X

2

_H

, where

G = 1

−

higher terms,

H

∈ C[X]

. Then

yg = x

2

_h

_∈

[F ]

so

y = x

2

_hg

−1

_{∈ (x)}

, sin e

g(P )

6= 0

. Thus

M

P

(F ) = (x, y) = (x)

as desired.

Asforthe onverse:if

O

P,F

isadis retevaluationring,thenbyTheorem(1.3.29) and Remark(1.3.27), it follows immediatelythat

m

P

(F ) = 1

.



Corollary 1.3.31 Let

P

be a point on an irredu ible urve

F

and

M =

M

p

(F )

. Thendim

C

(M

n

_/M

n+1

_{) = n + 1}

for

0

≤ n < m

p

(F )

. Inparti ular,

P

is asimple point ifand only ifdim

C

(M/M

2

_{) = 1}

;

other-wise dim

C

(M/M

2

_{) = 2}

.

Proof As in the Theorem(1.3.29), fromexa t sequen e

0

_−−−→

_M

M

n+1

n

−−−→

_M

O

n+1

−−−→

_M

O

n

−−−→ 0

itisenoughto al ulatedim

C

O

M

n+1

−

dim

C

(

O

M

n

)

.ButagainasinTheorem(1.3.29),

O

M

n

∼

= C[X, Y ]/(I

n

, F )

and sin e by hypothesis

n < m

P

(F )

it follows that

C[X, Y ]/(I

n

_{, F ) = C[X, Y ]/(I}

n

₎

, so by Remark(1.3.25) it follows

immedi-ately that dim

C

O

M

n+1

−

dim

C

(

O

M

n

) = n + 1



We introdu e the followingdenition.

Denition 1.3.32 (Interse tion Number) Let

F

and

G

be plane urves,

P

∈

C

2

. The number

i(P, F

_{∩ G) =}

dim

C



O

P,C

2

(F, G)



is alled the interse tionnumber of

F

and

G

at

P

.

Denition 1.3.33 (Transversal Interse tion) Two urves

F

and

G

are said to interse ttransversally at

P

if

P

isa simplepoint both on

F

and on

G

, and if the tangentline to

F

at

P

isdierent from the tangentline to

G

at

P

. Theorem 1.3.34 The interse tion number satisfy the followingproprieties:

(1) If

T

is an ane hange of oordinates on

C

2

, and

T (P ) = Q

, then

i(Q, F

T

_{∩ G}

T

) = i(P, F

_{∩ G).}

(2)

i(P, F

∩G) ≥ m

p

(F )m

P

(G)

,withequalityo urringifandonlyif

F

and

G

havenotangentlinesin ommonat

P

.Inparti ular,theinterse tion number is one when

F

and

G

meet transversally at

P

.

(3) If

F =

Q

F

r

i

i

, and

G =

Q

G

s

j

j

, then

i(P, F

∩ G) =

X

i,j

r

i

s

j

i(P, F

i

∩ G

j

).

Lemma 1.3.35 Let

F

and

G

be plane urves,

P

∈ C

2

and

m = m

P

(F ), n =

m

P

(G)

. Let

I = (X, Y )

⊂ C[X, Y ]

and

ψ :

C[X, Y ]

I

n

×

C[X, Y ]

I

m

−→

C[X, Y ]

I

n+m

the map given by

ψ(A, B) = AF + BG

. Then: (a) If

F

and

G

havedistin t tangentsat

P

, then

I

t

_{⊂ (F, G)O = O}

P,C

2

for

t

≥ m + n − 1

.

(b)

ψ

isone-to-one if and only if

F

and

G

have distin ttangents at

P

. Proof We mayassume

P = (0, 0)

,and that allthe omponents of

F

and

G

pass trough

P

.

(a)Let

L

1

, . . . , L

m

and

M

1

, . . . , M

n

be respe tively the tangentto

F

and

G

at

P

(note that sin e

F

and

G

have distin t tangents at

P

, it holds that no

L

i

= γM

j

,

γ

∈ C

). Let

L

i

= L

m

if

i > m

,

M

j

= M

n

if

j > n

, and let

A

ij

= L

1

· · · L

i

· M

1

· · · M

j

for all

i, j

≥ 0

(

A

00

= 1

). Then, sin e by Lemma(1.3.9)

{A

ij

|i + j = t}

forms a basis for the ve tor spa e of all forms of a degree

t

in

C[X, Y ]

, it is enough to show that

A

ij

∈ (F, G)O

for all

i + j

_{≥ m + n − 1}

.

Sin e

i + j

≥ m + n − 1

then either

i

≥ m

or

j

≥ n

. Say

i

≥ m

, so

A

ij

= A

m0

B

, where

B

is a form of degree

i + j

− m

. But, sin e

F

an be written as

F = A

m0

+ F

′

,where allterms of

F

′

are of degree

≥ m + 1

, then

A

ij

= BF

−BF

′

,whereea htermof

BF

′

hasdegree

≥ i+j +1

.Thenwe an do againthis pro ess for

BF

′

,so weare redu ed toprove that

I

t

_{⊂ (F, G)O}

for all su iently large

t

.

Let

V (F, G) =

{P, Q

1

, . . . , Q

s

}

, and hoose as in Remark(1.3.19) a poly-nomial

H

so that

H(Q

i

) = 0

,

H(P )

6= 0

. Sin e

HX, HY

∈ I(V (F, G))

, and by Nullstellensatz

I(V (F, G)) =

Rad

((F, G))

, then

(HX)

N

_{, (HY )}

N

_∈

(F, G)

⊂ C[X, Y ]

for some

N

. Finally, sin e

H

N

is a unit in

O

, it follows that

X

N

_{, Y}

N

_{∈ (F, G)O}

,and therefore

I

2N

_{⊂ (F, G)O}

as desired.

(b) Let

A = A

r

+

(higher terms) and

B = B

s

+

(higher terms), and sup-pose

ψ(A, B) = AF + BG = 0

. This means that

AF + BG = A

r

F

m

+

B

s

G

n

+

(higher terms) has only terms of degree

≥ m + n

; thus

A

r

F

m

=

−B

s

G

n

. But, sin e by hypothesis

F

and

G

have distin t tangents,

F

m

and

G

n

haveno ommonfa tors, so

F

m

divides

B

s

and

G

n

divides

A

r

.Therefore

s

_{≥ m, r ≥ n}

, so

(A, B) = (0, 0)

.

Conversely, if

L

was a ommon tangent to

F

and

G

at

P

, then

F

m

=

LF

′

m−1

, G

n

= LG

′

_n−1

.But then

ψ(G

′



Remark 1.3.36 Let

I

be an ideal in

C[X

1

, . . . , X

n

]

. Then if

V (I)

is an innite set then dim

C



_C

[X

1

,...,X

n

]

I



=

_∞

.

Proof Ifnot,forany

r

points

P

1

, . . . , P

r

∈ V (I)

,wehave

r

≤

dim

C



_C

[X

1

,...,X

n

]

I



.

Indeed, by Remark(1.3.19), we an hoose

F

i

. . . , F

r

∈ C[X

1

, . . . , X

n

]

su h that

F

i

(P

j

) = 0

if

i

6= j

, and

F

i

(P

i

) = 1

. Let

P

λ

i

F

i

= 0

,

λ

i

∈ C

, then

P

λ

i

F

i

∈ I

, so

λ

i

= (

P

λ

i

F

i

)(P

j

) = 0

; thus the

F

i

are linearly independent over

C

.



Proof (ofTheorem(1.3.34))(1)Itfollowsimmediatelyfromthefa tthat,

by Lemma(1.3.8), an ane hange of oordinates gives an isomorphism of

lo alrings.

Nowwe mayassume

P = (0, 0)

,and thatallthe omponentsof

F

and

G

pass trough

P

.

(2)Let

m = m

P

(F )

and

n = m

P

(G)

. Let

I = (X, Y )

⊂ C[X, Y ]

and

ψ :

C[X, Y ]

I

n

×

C[X, Y ]

I

m

−→

C[X, Y ]

I

n+m

the map given by

ψ(A, B) = AF + BG

. Then the sequen e

C

[X,Y ]

I

n

×

C

[X,Y ]

I

m

ψ

−−−→

C

I

[X,Y ]

n+m

ϕ

−−−→

(I

n+m

C

[X,Y ]

_,F,G)

−−−→ 0

where

ϕ

is the natural ring homomorphism, isexa t. Itfollows that dim



C[X, Y ]

I

n



+

dim



C[X, Y ]

I

m



≥

dim

(

Ker

(ϕ)) ,

with the equality if and onlyif

ψ

is one-to-one, and that dim



C[X, Y ]

(I

n+m

_{, F, G)}



=

dim



C[X, Y ]

I

n+m



−

dim

(

Ker

(ϕ)) .

Now note that

V (I

n+m

_{, F, G)}

_{⊂ {P }}

, so by Corollary(1.3.22), we have that

C

[X,Y ]

(I

n+m

_,F,G)

is isomorphi to

O

(I

n+m

_,F,G)

, where

O = O

P,C

2

. Finally, let

π :

O

(F,G)

→

(I

n+m

O

_,F,G)

be the natural ring homomorphism, that is learly sur-je tive.

From allthese onsiderationsplus Remark(1.3.25) we have

i(P, F

_{∩G) =}

dim



O

(F, G)

O



≥

dim



O

(I

n+m

_{, F, G)}

_O



=

dim



C[X, Y ]

(I

n+m

_{, F, G)}



≥