Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Containment Problems and the Resurgence for
Points on Intersecting Lines in P2
Introduction
Preliminaries
Main Results
Another
Approach
Annika Denkert
Joint work with M. Janssen
Department of Mathematics
October 15, 2011
1 / 15
The Question
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
Preliminaries
Main Results
Another
Approach
2 / 15
Given a homogeneous ideal I in a ring R, how do I (m) and I r
compare?
In particular, for which m and r do we have I (m) ⊆ I r ?
This question is still open in general. We will answer the
question for a specific ideal of points here.
Definition (Ideal, Symbolic Powers, and the Resurgence)
Let k be an infinite field, R = k[x0 , . . . , xN ], and
p 0 , . . . , p n ∈ PN
k distinct points.
Tn
Then the
ideal
of
the
points
is
I
=
i=0 I(pi ) and
Tn
(m)
m
I
= i=0 I(pi ) .
(m)
I
6⊆ I r .
Define the resurgence ρ of I by ρ(I) = sup m
r
Containment
Problem
I
In 2001, L. Ein, R. Lazarsfeld, and K. Smith proved that for
a large class of ideals, I (nc) ⊆ I n for all n ≥ 0, where c is
the largest height of an associated prime of I. M. Hochster
and C. Huneke have since expanded on those results.
I
More recently, C. Bocci and B. Harbourne gave a complete
answer for ideals defined by particular configurations of
points in P2 .
I
We will consider two “extremal” configurations of points
in P2 - one with equally many points on two intersecting
lines (this talk) and one with all points except one on the
same line (M. Janssen’s talk).
Annika
Denkert
Joint work
with
M. Janssen
Introduction
Preliminaries
Main Results
Another
Approach
3 / 15
Some Facts
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
Proposition
Let m, r ∈ N. Assume I ( k[PN ] is nontrivial.
m
r
I
I (m) ⊆ I r if
I
I (m) = I m if I is a complete intersection.
I
I r ⊆ I (m) if and only if m ≤ r.
I
I (m) ⊆ I r implies m ≥ r.
I
I (m) ⊆ I r if m ≥ N r.
> ρ(I) by definition of ρ(I).
Preliminaries
Main Results
Another
Approach
Since we know exactly when I r ⊆ I (m) , we want to find
stronger necessary and sufficient conditions for I (m) ⊆ I r than
we currently have.
4 / 15
The Set-Up
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
We will consider the following configuration of 2n + 1 points in
P2 .
z=0
y=0
P2
n points
Preliminaries
Main Results
Another
Approach
x=0
n points
Then I = (x, y) ∩ (xy, F ) = (xy, xF, yF ) for a form F of
degree n ≥ 1, and
I (m) = (x, y)(m) ∩ (xy, F )(m) = (x, y)m ∩ (xy, F )m as (x, y)
and (xy, F ) are complete intersections.
We can actually give F explicitly, but that isn’t necessary for the purpose of this talk.
5 / 15
A k-basis of R, I (m) , and I r
Containment
Problem
The set
Annika
Denkert
Joint work
with
M. Janssen
Lemma
xa y b z c F d c < n is a k-basis for R = k[x, y, z].
Let m ≥ 1. Then
1. (x, y)m = hxa y b z c F d |c < n, a + b ≥ mi
Introduction
Preliminaries
2. (xy, F )m = hxa y b z c F d |c < n, min(a, b) + d ≥ mi
Main Results
Another
Approach
3. I (m) = (x, y)m ∩ (xy, F )m
= hxa y b z c F d |c < n, a + b ≥ m, min(a, b) + d ≥ mi
Lemma
Let r ≥ 1. Then
I r = hxa y b z c F d |c < n, a+b ≥ r, min(a, b)+d ≥ r, a+b+d ≥ 2ri.
6 / 15
Recap
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Let m, r ≥ 1. Then
I (m) = hxa y b z c F d |c < n, a + b ≥ m, min(a, b) + d ≥ mi
and
Introduction
Preliminaries
I r = hxa y b z c F d |c < n, a+b ≥ r, min(a, b)+d ≥ r, a+b+d ≥ 2ri.
Main Results
Another
Approach
7 / 15
The conditions on (a, b, c, d) in the description of I r mirror
those given in the description of I (m) , except for that last
condition, a + b + d ≥ 2r. It turns out that this is the
condition that determines containment of I (m) in I r .
Containment and the Resurgence
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Theorem (1)
We have I (m) * I r if and only if
(a) either r > m or
Introduction
Preliminaries
Main Results
(b) r ≤ m and there exists xa y b F d ∈ I (m) such that
a + b + d < 2r.
Another
Approach
Theorem (2)
For m, r ≥ 1 we have I (m) * I r if and only if 4r > 3m + 1.
In particular, the resurgence ρ(I) is 43 .
8 / 15
Proof of (1) ⇒ (2)
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
(a) r > m ⇒ 4r > 4m ≥ 3m + 1
X
(b) r ≤ m, some xa y b F d ∈ I (m) with a + b + d < 2r:
4r > 2(a + b + d) ≥ 3m as
a+b≥m
a+d≥m
b+d≥m
Preliminaries
Main Results
Another
Approach
m odd:
4r > 2(a + b + d) ≥ 3m ⇔ 4r > 2(a + b + d) ≥ 3m + 1
X
m even:
4r > 3m ⇔ 2r > 3m/2
⇔ 2r ≥ 3m/2 + 1
⇔ 4r ≥ 3m + 2
⇔ 4r > 3m + 1
9 / 15
X
Conjectures
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Conjectures
Let R = k[PN ], M = (x0 , . . . , xN ) the irrelevant ideal, and
J ( R a nontrivial ideal. Then for all m, r ∈ N
I
(B. Harbourne, C. Huneke) If J is a radical ideal of a finite
set of points, then J (N r) ⊆ Mr(N −1) J r and
J (N r−N +1) ⊆ M(r−1)(N −1) J r .
I
(B. Harbourne, C. Huneke) If J is a radical ideal
r of a finite
set of points, then J (r(m+N −1)) ⊆ Mr J (m) .
I
(B. Harbourne) If J is homogeneous, then
J (N r−N +1) ⊆ J r .
Introduction
Preliminaries
Main Results
Another
Approach
10 / 15
These conjectures are true for our ideal I.
We can, in fact, give a complete answer to the question when
I (m) is contained in Mt I r for any t, r ∈ N0 .
I (m) ⊆ Mt I r
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
Preliminaries
Main Results
Another
Approach
11 / 15
Let M = (x, y, z).
To find out for which (m, t, r) we have I (m) ⊆ Mt I r , we need
to assume that I (m) ⊆ I r . Therefore, we need to assume that
4r ≤ 3m + 1.
Let rmax denote the largest r such that I (m) ⊆ I r , i.e.
rmax = m − d m−1
4 e.
We can write
Mt I r = hxa y b z c F d |c < n, ∃α ≤ a, β ≤ b, δ ≤ d 
(a − α) + (b − β) ≥ r, min(a − α, b − β) + (d − δ) ≥
r, (a − α) + (b − β) + (d − δ) ≥ 2r, and α + β + nδ ≥
max(0, t − c)i.
This description is helpful in proving the following theorem.
I (m) ⊆ Mt I r
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Theorem
Let m ∈ N.
If r = rmax , then
I
I (m) ⊆ I r but I (m) * MI r if m ≡ 0, 1 mod 4.
I
I (m) ⊆ Mt I r but I (m) * Mt+1 I r , where
t = min(n, 2m − 2r) if m ≡ 2, 3 mod 4.
Introduction
Preliminaries
Main Results
Another
Approach
If r < rmax , then
I r but I (m) * Md
3m
e−2r+1
2
I (m) ⊆ Md
I
I (m) ⊆ Mt I r but I (m) * Mt+1 I r , where
I
I
12 / 15
3m
e−2r
2
I
t = 2m − 2r if r ≤ b m
2 c and n ≥ 2.
m
t = n d 3m
2 e − 2r if r > b 2 c and n ≥ 2.
I r if n = 1.
Symbolic Powers as Ordinary Powers
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
Lemma
For any s, t ∈ N, we have
I
I (2st) = (I (2s) )t and
I
I (2s+t) = I (2s) I (t) .
Preliminaries
Main Results
Another
Approach
In fact, we can re-write all symbolic powers of I as follows.
Proposition
Let m ∈ N. Then
m
I I (m) = I (2) 2 if m is even.
I
13 / 15
I (m) = I (2)
m−1
2
I if m is odd.
I (m) and I r as sums of ideals
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
We can also write I (m) and I r as sums of ideals,
I
(m)
=
m
X
i=0
(xy)i (x, y)max(0,m−2i) F m−i
|
{z
}
Introduction
Preliminaries
Main Results
and
Ir =
Another
Approach
r
X
j=0
=Ji
(xy)j (x, y)r−j F r−j .
|
{z
}
=Kj
Theorem
For m, r ∈ N, we have I (m) ⊆ I r if and only if for all
0 ≤ i ≤ m there exists 0 ≤ j ≤ r such that Ji ⊆ Kj .
14 / 15
Where do we go from here?
Containment
Problem
Annika
Denkert
Joint work
with
M. Janssen
Introduction
We are trying to use this or the vector space approach to find
the resurgence in the following case.
z=0
Preliminaries
y=0
P2
Main Results
Another
Approach
n points
x=0
n points
15 / 15