Containment Problem for
points on another reducible conic
Mike Janssen
(joint with A. Denkert)
Department of Mathematics
October 16, 2011
[email protected]
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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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) = ∩i I (pi )m .
Example: If I defines a complete intersection, I (m) = I m .
Facts: Given 0 6= 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 ?
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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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:
n
o
ρ(I ) = sup m/r : I (m) 6⊆ I r
Obviously, m/r > ρ(I ) implies I (m) ⊆ I r .
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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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 6= 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.
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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Introduction
Notation
Points in P2 , ideals in R = k[P2 ] = k[x, y , z]:
y =0
x =0
p0
p1 p2 p3 · · · pn z = 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
Mike Janssen (UNL)
Containment Problem
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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 x e F i y j z l , where
0 ≤ e < n.
Idea of the proof: R = spank hx β y γ z δ i and deg F = n, so replace x bn with
F b.
Proposition (k-bases of (z, F )m and (x, y )m )
(a) The ideal (z, F )m is spanned by forms x e F i y j z l satisfying
e + in + ln ≥ mn for e, i, j, l ≥ 0.
(b) The ideal (x, y )m is spanned by forms x e F i y j z l satisfying
e + in + j ≥ m for e, i, j, l ≥ 0.
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Containment Problem
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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 x e F i y j z l , 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 x e F i y j z l 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) 6⊆ I 6 when n = deg F = 3.
Consider xF 2 z 5 ∈ 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 2 z 5 ∈
/ 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.
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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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) 6⊆ I r holds if and only if either
rn2 + rn − n − 2
m < n and m ≤
, or
n2
n2 r − n
.
m ≥ n and m ≤ 2
n −n+1
Theorem (Asymptotic Solution)
For the ideal I of n ≥ 3 collinear points and one point off the line,
ρ(I ) =
Mike Janssen (UNL)
n2
.
n2 − n + 1
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October 16, 2011
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Main Results
Computing ρ(I ) from the complete solution
Assume m ≥ n
n2 r − n
.
n2 − n + 1
m
n2
n2 − n/r
n2 r − n
Then
> 2
> 2
⇒m> 2
⇒ I (m) ⊆ I r .
r
n −n+1
n −n+1
n −n+1
n2
Thus, ρ(I ) ≤ 2
.
n −n+1
Conversely, take m = tn2 − 1 and r = t(n2 − n + 1).Then
n2 r − n
m≤ 2
, but
n −n+1
Recall I (m) 6⊆ I r ⇔ m ≤
n2
n2
m
tn2 − 1
=
,
so
ρ(I
)
≥
.
= lim
t→∞ r
t→∞ t(n2 − n + 1)
n2 − n + 1
n2 − n + 1
lim
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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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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October 16, 2011
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Applications
Two conjectures
Harbourne and Huneke conjectured:
Conjecture
I (2r ) ⊆ M r I r , where M is the ideal generated by the variables.
Conjecture
I (2r −1) ⊆ M r −1 I 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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Containment Problem
October 16, 2011
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Applications
Thank you!
Mike Janssen (UNL)
Containment Problem
October 16, 2011
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