Containment Problems and the Resurgence for Annika Denkert Points - - PowerPoint PPT Presentation

Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Containment Problems and the Resurgence for Points on Intersecting Lines in P2

Annika Denkert Joint work with M. Janssen

Department of Mathematics

October 15, 2011

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

The Question

Given a homogeneous ideal I in a ring R, how do I(m) and Ir compare? In particular, for which m and r do we have I(m) ⊆ Ir? 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 p0, . . . , pn ∈ PN

k distinct points.

Then the ideal of the points is I = n

i=0 I(pi) and

I(m) = n

i=0 I(pi)m.

Define the resurgence ρ of I by ρ(I) = sup m

r

• I(m) ⊆ Ir

.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

◮ In 2001, L. Ein, R. Lazarsfeld, and K. Smith proved that for

a large class of ideals, I(nc) ⊆ In 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.

◮ More recently, C. Bocci and B. Harbourne gave a complete

answer for ideals defined by particular configurations of points in P2.

◮ 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).

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Some Facts

Proposition

Let m, r ∈ N. Assume I k[PN] is nontrivial.

◮ I(m) ⊆ Ir if m r > ρ(I) by definition of ρ(I). ◮ I(m) = Im if I is a complete intersection. ◮ Ir ⊆ I(m) if and only if m ≤ r. ◮ I(m) ⊆ Ir implies m ≥ r. ◮ I(m) ⊆ Ir if m ≥ Nr.

Since we know exactly when Ir ⊆ I(m), we want to find stronger necessary and sufficient conditions for I(m) ⊆ Ir than we currently have.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

The Set-Up

We will consider the following configuration of 2n + 1 points in P2. n points x = 0 n points y = 0 z = 0 P2 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

Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

A k-basis of R, I(m), and Ir

The set

• xaybzcF d

c < n

• is a k-basis for R = k[x, y, z].

Lemma

Let m ≥ 1. Then

• 1. (x, y)m = xaybzcF d|c < n, a + b ≥ m
• 2. (xy, F)m = xaybzcF d|c < n, min(a, b) + d ≥ m
• 3. I(m) = (x, y)m ∩ (xy, F)m

= xaybzcF d|c < n, a + b ≥ m, min(a, b) + d ≥ m

Lemma

Let r ≥ 1. Then Ir = xaybzcF d|c < n, a+b ≥ r, min(a, b)+d ≥ r, a+b+d ≥ 2r.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Recap

Let m, r ≥ 1. Then I(m) = xaybzcF d|c < n, a + b ≥ m, min(a, b) + d ≥ m and Ir = xaybzcF d|c < n, a+b ≥ r, min(a, b)+d ≥ r, a+b+d ≥ 2r. The conditions on (a, b, c, d) in the description of Ir 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 Ir.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Containment and the Resurgence

Theorem (1)

We have I(m) Ir if and only if (a) either r > m or (b) r ≤ m and there exists xaybF d ∈ I(m) such that a + b + d < 2r.

Theorem (2)

For m, r ≥ 1 we have I(m) Ir if and only if 4r > 3m + 1. In particular, the resurgence ρ(I) is 4

3.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Proof of (1) ⇒ (2)

(a) r > m ⇒ 4r > 4m ≥ 3m + 1

• (b) r ≤ m, some xaybF d ∈ I(m) with a + b + d < 2r:

4r > 2(a + b + d) ≥ 3m as a + b ≥ m a + d ≥ m b + d ≥ m m odd: 4r > 2(a + b + d) ≥ 3m ⇔ 4r > 2(a + b + d) ≥ 3m + 1

• m even:

4r > 3m ⇔ 2r > 3m/2 ⇔ 2r ≥ 3m/2 + 1 ⇔ 4r ≥ 3m + 2 ⇔ 4r > 3m + 1

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Conjectures

Conjectures

Let R = k[PN], M = (x0, . . . , xN) the irrelevant ideal, and J R a nontrivial ideal. Then for all m, r ∈ N

◮ (B. Harbourne, C. Huneke) If J is a radical ideal of a finite

set of points, then J(Nr) ⊆ Mr(N−1)Jr and J(Nr−N+1) ⊆ M(r−1)(N−1)Jr.

◮ (B. Harbourne, C. Huneke) If J is a radical ideal of a finite

set of points, then J(r(m+N−1)) ⊆ Mr J(m)r.

◮ (B. Harbourne) If J is homogeneous, then

J(Nr−N+1) ⊆ Jr. 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 MtIr for any t, r ∈ N0.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

I(m) ⊆ MtIr

Let M = (x, y, z). To find out for which (m, t, r) we have I(m) ⊆ MtIr, we need to assume that I(m) ⊆ Ir. Therefore, we need to assume that 4r ≤ 3m + 1. Let rmax denote the largest r such that I(m) ⊆ Ir, i.e. rmax = m − ⌈ m−1

4 ⌉.

We can write MtIr = xaybzcF 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). This description is helpful in proving the following theorem.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

I(m) ⊆ MtIr

Theorem

Let m ∈ N. If r = rmax, then

◮ I(m) ⊆ Ir but I(m) MIr if m ≡ 0, 1 mod 4. ◮ I(m) ⊆ MtIr but I(m) Mt+1Ir, where

t = min(n, 2m − 2r) if m ≡ 2, 3 mod 4. If r < rmax, then

◮ I(m) ⊆ M⌈ 3m

2 ⌉−2rIr but I(m) M⌈ 3m 2 ⌉−2r+1Ir if n = 1.

◮ I(m) ⊆ MtIr but I(m) Mt+1Ir, where

◮ t = 2m − 2r if r ≤ ⌊ m

2 ⌋ and n ≥ 2.

◮ t = n

• ⌈ 3m

2 ⌉ − 2r

• if r > ⌊ m

2 ⌋ and n ≥ 2.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Symbolic Powers as Ordinary Powers

Lemma

For any s, t ∈ N, we have

◮ I(2st) = (I(2s))t and ◮ I(2s+t) = I(2s)I(t).

In fact, we can re-write all symbolic powers of I as follows.

Proposition

Let m ∈ N. Then

◮ I(m) =

• I(2) m

2 if m is even.

◮ I(m) =

• I(2) m−1

2

I if m is odd.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

I(m) and Ir as sums of ideals

We can also write I(m) and Ir as sums of ideals, I(m) =

m

• i=0

(xy)i(x, y)max(0,m−2i)F m−i

• =Ji

and Ir =

r

• j=0

(xy)j(x, y)r−jF r−j

• =Kj

.

Theorem

For m, r ∈ N, we have I(m) ⊆ Ir if and only if for all 0 ≤ i ≤ m there exists 0 ≤ j ≤ r such that Ji ⊆ Kj.

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Containment Problem Annika Denkert Joint work with

• M. Janssen

Introduction Preliminaries Main Results Another Approach

Where do we go from here?

We are trying to use this or the vector space approach to find the resurgence in the following case. n points x = 0 n points y = 0 z = 0 P2

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