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By the Nullstellensatz we have fi x y. If 2 I. If X o y o or 7 0. p Pz Ps. R yz XZ x. If X Z l o 7 x co or 2

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(1)

I R yz

XZ x

we use

the Nullshelleusatz and

consider

2 I

I

X2 yz

If X Z l o 7 x co or 2

If

X o y o or 7 0

If

2 I

y

X

By

the

Nullstellensatz

we

have

fI

x

y

n X

2

n z

l y x2

p Pz Ps

(2)

of Al At 2 bereft

k

hxiy.it

KID

xp t

foe XZ

y

y s

t4

2 t

f y

3 belong

to I Ily

FE XI E f fof fz EI

Conversely

let FEI

Introduce a grading on

Kay

t

kg

deg x 3 elegy_4

degz 5

fo f fz

homogeneous

of

degree

8,9 10

Modulo

J

we

may

write

any fEk xy 27

as

f limit AIN

t

BHI y

x z

plugin

x

I

3

y Atty

t

Blt

3

t4 at Its

o

z

t5 T T T

0 mod

3 I

mod 3 2 mod

3

A B C

o as polynomials wi x

f

o

J ZI

I cannot be

generated

by fewer elements

there

are no

elements of degrees

a

7

(3)

F x

yw

o contains theline

L

2

How

G

Xy Ew 0

Let c 2 1

2 iWyw

xy

zu y2nxz7 uh

3 YI

Claim

7

F

G C

v

L

11

lis L

Z

Clear since

F G

E

ICC

n X w

C

Let PE

2

EG p

c y z w

p

luh Uu v

u3 f

w o Hur x Orr

p

E L

I

f

wa o then

why

w

y

2

2

Xy x3

p

x X

x3

1 E

C

Note yw

ay

Ew Xiu n x yw 2w Xy xz

y2

0

3 closed

Alternation P'xp IP embedding

doily

Volf H novo

44

wok

Niko

4 1

2

yw uiuo lu.yku.ro

Vo Uf ro

u

y

(4)

a

Krull's

Principal ideal theorem

b

on

c

Ily

minors

of 1 II I

is

and

redundantno minor

Y 2 F G F

Xo

Xz

x

G Xz

XzpXz

XE Xz

Xo

Xz Xix

VEG

Contains

IX Xz xET X F iz G

Xoxz

X

Kal X F ko G

Feat I

since

I is prime

(5)

a

Let p

e C

CP2 why PE

Deetz

AT

ng

p

n

17 12

7 is either lil

y

X 0 or Iii

Xy

1 0

i

Oc.pt kEyI7

z mp

kEMcx.a

normal

Ocp C y tix

x

D

ex a normal

b wlog p Il

un vi w

0cgp k 4YwI KEV

w is

normal

Uavw

QQz p Hurry

2

kEuw

or

k Luiu.us ELE w3 ronal

(6)

c

bix y HEHE not normal

as

te KE

Es

but t

satisfies

XZ t2

o

d

A AIX

integral down

A

integrally closed

A

nomad

Ap

norml

up

Am und

tem

Ox

p

nonal

X nonal

e See

the

notes

(7)

normal noethian

al

Hartog's

Hmi R

integral domain

p R KIRI

R

P

htp P

C

we

want to

show

that f

extends

to

a

regular function

on

p

n

may

assume

Y

is

affine

Let A ACY Being normal

is a

local

property

after replacing

Y with

an open set may

assume Let

p

be a ht 1 prime so

p Mp

Y

is homal

Pick

gemp p Agc Ap

Y

p

U D

h

f

regular on

Dcg

hempD g

T

appears n

f

E

O

Y Dcg

Ag

here

f

E

AP

Hence

f EA by

Hartog

f

extends in a

nbh of p

ok

I is regular

on

Alt

o

but

there

is no extension to

Atl

(8)

Moy X and Y

are

affine

Let

ACH

K YI

n

Ox

p HEY

YI Hu

Yanks up

Oy q 4th

Ym

mg Given

of

Huh Yanks

mg5 Kai HI

mp

Define

f U Y

7

by f Hi int

9lydlxi xn7

199m74 xn

This

is defied in a ubha

of p

where

cely

oneregular

clearly

g up my

so

f Cpl of

If y p

we also get a map

g V

1

X glyi ym7 luckily

ym

which

is

defied

in a mbh

of g

we

have fog

id on

V

and

g of

id on

U

ok

(9)

See the notes on birational

maps

(10)

J

2

4

yup

to

to

2 4 2

E

90 only y Coco

Suiyuanpoint

Themultiplicity is 2 at p seria

tem

m

The

curve is symmetric iv

rat

x x

y y

tacnode

A B C D

f Hey

6

Xy

y

6 5

y 6y5

X

x6ty6 Xy xy

6y5_x 6 5

y

4xy

61 4

6 xy 1

16

5 y y 6y5 x

we have x o y o so 190 is the only singular

point

f invariant under

x 7 x

y

i

y

NODE

(11)

f y2xx4ty4 x3

I

J 4

3 3 2

4y3

12g

TI XT

4

37

0

II y

442 2 0

yao

y't y4

o KIu y o 0,0

4 0 x x3 o

IF

so Colo

11 34 and y I

t.i

does not lie on 2CI

0,07 is the only

super

point

This is thecusp

X3 y2

1 4

4430

no negative values

for

IIR

f X4ty4 Ry

Xyz

I

43 2

y y2

443 F 2

4

4 44

Ry

Xy

Macaulay 2

TI

Xy 907 only singularpoint

(12)

This has to be the triple pout by elevation It is clear

that p

coo7 has multiplicity 3

f x4ty4 Ry

Xyz

degreey degree3

If

f at bt7 is my line thogh p

thin

flat

btl l

It't f It't

(13)

A B

C

an F Xy 72 J 42 2xy ZZ

i suiylxl

ZLT.tl

ZLy z computeralong the

line y Z o

also X is

a fortyof

parabolas

b F

Rey

z J 2x 2y Zz

singlx a

2Gt

LED a 490,07 7

A

fairly

of

circles

c F

x3ty3

Xy

su

ylxt

ZCI lf.FI Zl3x2 y 3y2 x x3ty3 xy

Z does not appear air F

T

(14)

minyanim

ay yep

141

1

f axtby

1

g gadgeteer

F

xeaag.IOy

brfCo7

on

gem

p

non singular

b

a Hp Z 1710,07 become of the XZ

b

i

Yip

2 PEON become of

xp

c

Yp

2 at f to07

l

Il

ya

d Yp

3

at p 1907 My yt

(15)

Y t

district

Y

it is o dimensional

by Krull's PID

dim

1 integraldomain

Cf.gg 0440g artinian k

algebra non zerodivisor

Y Z

p dimk

YiP1g1Lx.b

Wl og p

Co

ol

c

AT

f f

a

fat

t Yipl

YI

a

G Cfb type

t Yipth b

R kGiy

µ hiy

y zip

dim

g

µ dimkff.mg

µatb

him

ftp.matbykoszul

complex

121µA X

RI yb R µatb Mf g µatb

0

UN 1 1

f

ut

g

v

dim Rdf

gµatby 7 dinis

Keat

b din14µA duiR

µb

atb

1

att

b

I

ab

(16)

C wlog p 10107

f Eaijx

yd

fat fat

1

f

Zly

ax

YplYl

d

Hasan Hi.Y

KEN

f

dlx ax fetchax t

For most aek

fda

ax to In that case

dimk

xy

fdxaxltf.ee dim k

1

d

c xd1

d L Ya

E limp

wlog Y

Hy

and all interaction points LnY lie inA

flxiyl

gcxl y hlx.gl

e e l

x

tag f

9 p dim kgFYY.ly

p matp Cxio7

zeroofmultm

L.y

ELL

YI e

PELnY

The point at infinity p Coilo

Y is given by 2

dgl

t y saz

et

4 copO d e n L Y d e te d

(17)

r

p char k

If

pt

d Ferment curve Xd t

ydt zd

0 works

J

died

dyd dzd

0

If pld Flay

z Xyd t

yzd

I z d I

2

yd

I 1 z

xd

2

yd

z d 2 efc

J

yall 2

xd

2

zd

t X

yd

2 d l y

za

2

Can check that 2 J n Y I set 7 1 etc

(18)

a Noles

b Why the Turgut lines are 11 0 any y O

flag _Xy thxy h highdegreeterns

Blow up U X Uoy o 40 1 chat y U X

Total tmfm

f

Cx4 7 x'u hlx.mx

X4u

1 h x und

F

Edith'Cxc4X

ailersects E 2CX at the point u x o

one morepoint

the other chart 4 1 X upy at

fluoy y y2uothluoy.gg y2fuo 0 4 0

(19)

UT is nonsrnynlw along E nY amice it is given

by

4 1 h x mx o and wot there

and there the 252 do not vanish

References

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