I R yz
XZ xwe use
the Nullshelleusatz and
consider2 I
I
X2 yz
If X Z l o 7 x co or 2
If
X o y o or 7 0If
2 Iy
XBy
theNullstellensatz
wehave
fI
xy
n X2
n zl y x2
p Pz Ps
of Al At 2 bereft
k
hxiy.itKID
xp t
foe XZ
y
y s
t4
2 t
f y
3 belongto I Ily
FE XI E f fof fz EI
Conversely
let FEI
Introduce a grading onKay
tkg
deg x 3 elegy_4
degz 5
fo f fz
homogeneousof
degree8,9 10
Modulo
J
wemay
writeany fEk xy 27
asf limit AIN
tBHI y
x zplugin
xI
3y Atty
tBlt
3t4 at Its
oz
t5 T T T
0 mod
3 I
mod 3 2 mod3
A B C
o as polynomials wi xf
oJ ZI
I cannot be
generatedby fewer elements
there
are noelements of degrees
a7
F x
yw
o contains thelineL
2How
G
Xy Ew 0Let c 2 1
2 iWywxy
zu y2nxz7 uh3 YI
Claim
7
FG C
vL
11lis L
Z
Clear sinceF G
EICC
n X wC
Let PE
2EG p
c y z wp
luh Uu vu3 f
w o Hur x Orrp
E LI
f
wa o thenwhy
wy
22
Xy x3
p
x Xx3
1 EC
Note yw
ay
Ew Xiu n x yw 2w Xy xzy2
0
3 closedAlternation P'xp IP embedding
doily
Volf H novo44
wokNiko
4 1
2yw uiuo lu.yku.ro
Vo Uf ro
uy
a
Krull's
Principal ideal theoremb
onc
Ily
minorsof 1 II I
isand
redundantno minorY 2 F G F
XoXz
xG Xz
XzpXzXE Xz
XoXz Xix
VEG
ContainsIX Xz xET X F iz G
Xoxz
XKal X F ko G
Feat I
sinceI is prime
a
Let p
e CCP2 why PE
DeetzAT
ng
p
n17 12
7 is either lily
X 0 or IiiXy
1 0i
Oc.pt kEyI7
z mpkEMcx.a
normalOcp C y tix
xD
ex a normalb wlog p Il
un vi w0cgp k 4YwI KEV
w isnormal
Uavw
QQz p Hurry
2
kEuw
or
k Luiu.us ELE w3 ronal
c
bix y HEHE not normal
aste KE
Esbut t
satisfiesXZ t2
od
A AIX
integral downA
integrally closedA
nomadAp
normlup
Am und
temOx
pnonal
X nonal
e See
the
notes
normal noethian
al
Hartog'sHmi R
integral domainp R KIRI
R
Phtp P
C
we
want to
showthat f
extendsto
aregular function
onp
n
may
assumeY
isaffine
Let A ACY Being normal
is alocal
propertyafter replacing
Y with
an open set mayassume Let
p
be a ht 1 prime sop Mp
Y
is homalPick
gemp p Agc Ap
Y
pU D
hf
regular onDcg
hempD g
T
appears nf
EO
Y DcgAg
here
f
EAP
Hence
f EA by
Hartogf
extends in anbh of p
okI is regular
onAlt
obut
thereis no extension to
Atl
Moy X and Y
areaffine
Let
ACHK YI
nOx
p HEYYI Hu
Yanks upOy q 4th
Ymmg Given
of
Huh Yanksmg5 Kai HI
mpDefine
f U Y
7
by f Hi int
9lydlxi xn7199m74 xn
This
is defied in a ubhaof p
wherecely
oneregularclearly
g up my
sof Cpl of
If y p
we also get a mapg V
1X glyi ym7 luckily
ymwhich
isdefied
in a mbhof g
we
have fog
id onV
andg of
id onU
okSee the notes on birational
maps
J
24
yupto
to2 4 2
E
90 only y Coco
Suiyuanpoint
Themultiplicity is 2 at p seria
tem
mThe
curve is symmetric ivrat
x xy y
tacnodeA B C D
f Hey
6Xy
y
6 5y 6y5
Xx6ty6 Xy xy
6y5_x 6 5y
4xy
61 4
6 xy 116
5 y y 6y5 xwe have x o y o so 190 is the only singular
point
f invariant under
x 7 xy
iy
NODEf y2xx4ty4 x3
IJ 4
3 3 24y3
12gTI XT
437
0II y
442 2 0yao
y't y4
o KIu y o 0,04 0 x x3 o
IF
so Colo11 34 and y I
t.i
does not lie on 2CI0,07 is the only
super
pointThis is thecusp
X3 y2
1 44430
no negative values
for
IIRf X4ty4 Ry
XyzI
43 2y y2
443 F 24
4 44Ry
XyMacaulay 2
TI
Xy 907 only singularpointThis has to be the triple pout by elevation It is clear
that p
coo7 has multiplicity 3f x4ty4 Ry
Xyzdegreey degree3
If
f at bt7 is my line thogh pthin
flat
btl lIt't f It't
A B
C
an F Xy 72 J 42 2xy ZZ
i suiylxl
ZLT.tl
ZLy z computeralong theline y Z o
also X is
a fortyof
parabolas
b F
Rey
z J 2x 2y Zzsinglx a
2Gt
LED a 490,07 7A
fairlyof
circlesc F
x3ty3
Xysu
ylxt
ZCI lf.FI Zl3x2 y 3y2 x x3ty3 xyZ does not appear air F
T
minyanim
ay yep
141
1f axtby
1g gadgeteer
F
xeaag.IOybrfCo7
ongem
p
non singularb
a Hp Z 1710,07 become of the XZ
b
iYip
2 PEON become ofxp
c
Yp
2 at f to07l
Ilya
d Yp
3at p 1907 My yt
Y t
districtY
it is o dimensionalby Krull's PID
dim
1 integraldomainCf.gg 0440g artinian k
algebra non zerodivisorY Z
p dimk
YiP1g1Lx.b
Wl og p
Cool
cAT
f f
afat
t YiplYI
aG Cfb type
t Yipth bR kGiy
µ hiy
y zip
dimg
µ dimkff.mgµatb
himftp.matbykoszul
complex121µA X
RI yb R µatb Mf g µatb
0UN 1 1
f
utg
vdim Rdf
gµatby 7 dinis
Keat
b din14µA duiRµb
atb
1att
bI
abC wlog p 10107
f Eaijx
ydfat fat
1f
Zly
axYplYl
dHasan Hi.Y
KEN
f
dlx ax fetchax tFor most aek
fda
ax to In that casedimk
xyfdxaxltf.ee dim k
1
dc xd1
d L Ya
E limp
wlog Y
Hy
and all interaction points LnY lie inAflxiyl
gcxl y hlx.gle e l
x
tag f
9 p dim kgFYY.lyp matp Cxio7
zeroofmultm
L.y
ELL
YI ePELnY
The point at infinity p Coilo
Y is given by 2
dgl
t y sazet
4 copO d e n L Y d e te d
r
p char k
If
pt
d Ferment curve Xd tydt zd
0 worksJ
dieddyd dzd
0If pld Flay
z Xyd tyzd
I z d I2
yd
I 1 zxd
2yd
z d 2 efcJ
yall 2xd
2zd
t Xyd
2 d l yza
2Can check that 2 J n Y I set 7 1 etc
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.mxX4u
1 h x undF
Edith'Cxc4Xailersects 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
UT is nonsrnynlw along E nY amice it is given
by
4 1 h x mx o and wot thereand there the 252 do not vanish