Containment Problem for points on another reducible conic Mike - - PowerPoint PPT Presentation

Containment Problem for points on another reducible conic

Mike Janssen (joint with A. Denkert)

Department of Mathematics

October 16, 2011 s-mjansse7@math.unl.edu

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Introduction

The Problem

Recall: Q: Given an ideal I, how do I (m) and I r compare? If I ⊆ R = k[PN] is the ideal of points p1, p2, . . . , pd ∈ PN, then I (m) = ∩iI(pi)m. Example: If I defines a complete intersection, I (m) = I m. Facts: Given 0 = I R = k[PN] homogeneous, I r ⊆ I (m) ⇔ r ≥ m. I (m) ⊆ I r ⇒ m ≥ r, so assume m ≥ r. Containment Problem (CP): For which m and r is I (m) ⊆ I r?

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Introduction

Two ways to solve the CP

(Exact Solution) Find the set of all (m, r) such that I (m) ⊆ I r. (Asymptotic Solution) Find the resurgence, ρ(I), defined by Bocci and Harbourne: ρ(I) = sup

• m/r : I (m) ⊆ I r

Obviously, m/r > ρ(I) implies I (m) ⊆ I r.

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Introduction

Some facts about ρ(I)

Theorem (Hochster-Huneke) For I ⊆ k[PN] homogeneous, I (rN) ⊆ I r. Corollary If I ⊆ k[PN] is nontrivial and homogeneous, 1 ≤ ρ(I) ≤ N. For ideals 0 = I k[PN] homogeneous, If I defines a complete intersection, then ρ(I) = 1. No I is known with ρ(I) = N. Computing ρ is hard; complete solutions are even harder. Exact values of ρ are known in only a few cases. Today: I defines points on a pair of lines.

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Introduction

Notation

Points in P2, ideals in R = k[P2] = k[x, y, z]: p1 p2 p3 · · · p0 pn z = 0 x = 0 y = 0 n ≥ 3 I(p0) = (x, y) and I(p1 + · · · + pn) = (z, F), F ∈ k[x, y], deg F = n I = I(p0 + p1 + · · · + pn) = (x, y) ∩ (z, F) = (xz, yz, F) I (m) = (x, y)(m) ∩ (z, F)(m) = (x, y)m ∩ (z, F)m

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Preliminaries

The idea

Find compatible k-bases of the ideals

Theorem (k-basis of R) R = k[x, y, z] is spanned by “monomials” of the form xeF iyjzl, where 0 ≤ e < n. Idea of the proof: R = spankxβyγzδ and deg F = n, so replace xbn with F b. Proposition (k-bases of (z, F)m and (x, y)m) (a) The ideal (z, F)m is spanned by forms xeF iyjzl satisfying e + in + ln ≥ mn for e, i, j, l ≥ 0. (b) The ideal (x, y)m is spanned by forms xeF iyjzl satisfying e + in + j ≥ m for e, i, j, l ≥ 0.

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Preliminaries

k-bases of I (m) and I r

Corollary (k-basis of I (m)) Let m ≥ 1. Recall I (m) = (x, y)m ∩ (z, F)m. Then I (m) is spanned by “monomials” of the form xeF iyjzl, where e, i, j, l ≥ 0, 0 ≤ e < n and (a) e + in + ln ≥ mn, and (b) e + in + j ≥ m. Proposition (k-basis of I r) Let r ≥ 1. Then I r is spanned by elements of the form xeF iyjzl with e, i, j, l ≥ 0 and: (a) l < j and e + in + nl ≥ rn, or (b) e + in + j ≤ l and e + in + j ≥ r, or (c) j ≤ l < e + in + j and e + in + j + (n − 1)l ≥ rn. (c) j ≤ l < e + in + j and e + in + j + (n − 1)l ≥ rn.

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Preliminaries

An Example

Claim: I (7) ⊆ I 6 when n = deg F = 3. Consider xF 2z5 ∈ I (7) since it satisfies the inequalities (a) e + in + ln ≥ mn (i.e., 1 + 2 · 3 + 5 · 3 ≥ 7 · 3), and (b) e + in + j ≥ m (i.e., 1 + 2 · 3 + 0 ≥ 7) but xF 2z5 / ∈ I 6 since j ≤ l < e + in + j (i.e., 0 ≤ 5 < 1 + 2 · 3 + 0) but e + in + j + (n − 1)l ≥ rn (i.e.,1 + 2 · 3 + 0 + (3 − 1) · 5 = 17 ≥ 18 = 6 · 3) fails.

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Main Results

Two solutions to the CP

Theorem (Complete Solution) Let I be the ideal of n ≥ 3 collinear points and one point off the line. Then I (m) ⊆ I r holds if and only if either m < n and m ≤ rn2 + rn − n − 2 n2 , or m ≥ n and m ≤ n2r − n n2 − n + 1. Theorem (Asymptotic Solution) For the ideal I of n ≥ 3 collinear points and one point off the line, ρ(I) = n2 n2 − n + 1.

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Main Results

Computing ρ(I) from the complete solution

Assume m ≥ n

Recall I (m) ⊆ I r ⇔ m ≤ n2r − n n2 − n + 1. Then m r > n2 n2 − n + 1 > n2 − n/r n2 − n + 1 ⇒ m > n2r − n n2 − n + 1 ⇒ I (m) ⊆ I r. Thus, ρ(I) ≤ n2 n2 − n + 1. Conversely, take m = tn2 − 1 and r = t(n2 − n + 1).Then m ≤ n2r − n n2 − n + 1, but lim

t→∞

m r = lim

t→∞

tn2 − 1 t(n2 − n + 1) = n2 n2 − n + 1, so ρ(I) ≥ n2 n2 − n + 1.

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Applications

Symbolic powers are ordinary powers

Theorem If I is the ideal of n collinear points and one point off the line, then I (nt) = (I (n))t for all t ≥ 1. Moreover, n is the least positive integer for which this equality holds for all t ≥ 1. As a consequence, the symbolic power algebra ⊕I (m) is Noetherian.

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Applications

Two conjectures

Harbourne and Huneke conjectured: Conjecture I (2r) ⊆ MrI r, where M is the ideal generated by the variables. Conjecture I (2r−1) ⊆ Mr−1I r, where M is the ideal generated by the variables. Both are true for the ideal I of n collinear points and one point off the line.

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Applications

Thank you!

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