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)R regular ring I radical ordeal

Iprime ideal

h - big height of I = Max }htQ : Qetssct ) }then - th symbolic power of the Prime

I is

It '=I "

RIAR

=3fer : GFEI"

for some g¢I}={ fer : gtepnrp for some get I }=I - primarycomponent in an irredundant primary decomposition of I

"

= functions that vanish up to adorn in the variety defined byI .

IN = n #Rank )QEMINCI)

Iopertus • In c- In 'symbolic powers

are hard to study.

-

• tennis ten ) Even determining a minimal setof• In # Ian in general generators for Ian

,

a what degreesthey hue in

,can be very hand .

Example

I=Ker(K[x ,y,z]→K[t3,t4,t5 ] ) degx =3

-

= ( x3 .

yz , ya - xz, z2-x2y ) degy -4

hf wgn Tf degz -5

dg 9 deg8 deg 10

I22f£gh=qx , x¢I ⇒ QEIH ⇒ dgq = 15

.

aegisTG3 But all elements in # have degree > 16 ⇒ q¢I2

.

-

.I2 EIQ )

Tnnfaots : a) In # Ia)

for all n 2) IN e Pt

Continuation (sekenzeflasos) when Is Idle Ib ?

them CEinthazasfeld- South,door

,Holist - Haneke

,2002

,ha - Schwede

,207 )

at I be a radical idealof big height h in a regular local ring R

.

then

tchn )a- In for all n >- 1

.

Clouston (Hunekgaooo) Ipime of

height2 in a RLR

.

Does that imply PB ) c- I2 ?

Conjecture ( Harbaoone ,s 2008 ) I radical ideal of big height h in a regular ring-

Ichn- ht ' )

c- In.

Example ( Demnicki,Stemberg ,Tutaj - Gasinbka

,2013 ) char k¥2,3

I = ( x ( ya. za ), ZCE- na) , 2-Getya )) E K[x,g,z ]

= Ia ( graft8nF Efg ) Its '¢ It

But ! this example Is not prime. Huneke's question remains open .

In fact , Harbourne 's Conjecture holds for :

• points in general position in P2( Harbaoehe . Huneke ) and P3( Dumnicki )

• monomial ideals

• ideals defining I -

pure rings ( G - Huneke ) meaning ,

PYIF- pure .

From now on : R=

ktxy,

ZI,

K a field .

Opd Clifton ( Huneke ,2000 ) I prime of height2. Is I

" 'eI2 ?

them f) char

k¥3,I= kodpihsysz ] → Kita

, tbstc ] ) .

Faotcltrzoy ) I=±a(z£3xIIyFI)

sostudyhtidabeqwatnateftoam±= to ( [ ah agnag,

) ) ai,bier

.

he will follow Alexandra sealeanu 's work - she found conditions implying±3¥I?

But we will use the same ingredients to obtain IAIIB.beginsIn

'= In

:m° for all n > 1

Consequence:

- Ica )c- Ib

⇐ H° ( R/±a ) Is H° ( R/±b )m m

⇐ Ext's C RH, R ) -0 Ext3(R1Ib

,

R ) ( local duality )

⇐ EXTTTIR) °→Ext2C±gR ) ( Ext shifting )map induced by IAEIB

one-Man: Find resolutions to all In,

then determine lefts for Intte In :

- In → o Apply HOMRCTR )° →

Fy→

If→

¥0'→g±m → o

Compute homology .

0 → G → Gz → Goa

How do we do this ? we use Rees Algebras !

the Rees Algebra of I is the graded algebra +0 Int"

E RE t ]

there Is a graded map R[ I, I , TD - ⇒RCI )

Ti - fit

In general , determining the kernel of this map is hardy but our setting Is nice :

RCI ) ± REE ,IF3]/(qt±+ astataztz ,bet best bps )

¥ -

G

np Since RCI ) Is a complete intersection, the koszul complex gives a

free resolution for RCI ) Over FREI, I ,T3] :

0 → SC-2) - SG ) @ S C- i ) - S - RCI ) → 0

( E ] [ G - FI

Sn = R } monomials in Esta ,t3 of degree n } ±R(" ¥ )

o →

penal→ pt£ ) @ pf "I ) -

RC "2 )- In → o

in on

# n£3) → In 't→ o

Now how do we left Inttc In ?

D= f±%t±+fa%tatf3%I Euler operator

Induces a map on RCI ) of degree -1 corresponding to

n ( the Int ) in degree n : g#¥"

INETGYIII

Example : It 3) E Is,

char k¥3

O ← R ← 123 @ 123 c- 126 ← It ← 0

c t t t t3 10

-

o ← R ← 1260126 ← R ← Is ← 0

E

to 'e±a ⇐ Hateimfootaoaobaogkgnaghoobtoebqsoaobqoob )fzC

E

theorem ( G -

Hunekeflukundan) If µ( agog

,as

,by ba.bz ) = 5

,

then It 3) E I2.

exampt EstatesIt Ej'D%naEI¥FE,

Fatigate 't ..

then G) If D= kerfktx, zz )

→ k[t9tbt9 ),

ICIKP ?as long as char k¥5,2 .

these G) If I=koe(k[By ,z] → k[t9t3t9 )for a =3 a4 £ be C

,then I 4)

E

Itfoechaekta.

Example: when I = ker(k[ By , z ]→ k[t9t",t' 4 ] )

,

we have I4 ' ¢I3 . However,Ian - 1)

c- In Vn > a.

( In fact , for n > 8,

Plan - 2)c- In )