Containment Problem Containment Problems and the Resurgence for Annika Denkert Points on Intersecting Lines in P 2 Joint work with M. Janssen Introduction Preliminaries Annika Denkert Main Results Joint work with M. Janssen Another Approach Department of Mathematics October 15, 2011 1 / 15
The Question Containment Given a homogeneous ideal I in a ring R , how do I ( m ) and I r Problem Annika compare? Denkert In particular, for which m and r do we have I ( m ) ⊆ I r ? Joint work with M. Janssen This question is still open in general. We will answer the Introduction question for a specific ideal of points here. Preliminaries Main Results Definition (Ideal, Symbolic Powers, and the Resurgence) Another Let k be an infinite field, R = k [ x 0 , . . . , x N ] , and Approach p 0 , . . . , p n ∈ P N k distinct points. Then the ideal of the points is I = � n i =0 I ( p i ) and I ( m ) = � n i =0 I ( p i ) m . � m � � I ( m ) �⊆ I r � Define the resurgence ρ of I by ρ ( I ) = sup . r 2 / 15
Containment Problem ◮ In 2001, L. Ein, R. Lazarsfeld, and K. Smith proved that for Annika a large class of ideals, I ( nc ) ⊆ I n for all n ≥ 0 , where c is Denkert Joint work the largest height of an associated prime of I . M. Hochster with M. Janssen and C. Huneke have since expanded on those results. Introduction Preliminaries ◮ More recently, C. Bocci and B. Harbourne gave a complete Main Results answer for ideals defined by particular configurations of Another points in P 2 . Approach ◮ We will consider two “extremal” configurations of points in P 2 - 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). 3 / 15
Some Facts Containment Proposition Problem Let m, r ∈ N . Assume I � k [ P N ] is nontrivial. Annika Denkert ◮ I ( m ) ⊆ I r if m Joint work r > ρ ( I ) by definition of ρ ( I ) . with M. Janssen ◮ I ( m ) = I m if I is a complete intersection. Introduction Preliminaries ◮ I r ⊆ I ( m ) if and only if m ≤ r . Main Results Another ◮ I ( m ) ⊆ I r implies m ≥ r . Approach ◮ I ( m ) ⊆ I r if m ≥ Nr . 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 We will consider the following configuration of 2 n + 1 points in Containment P 2 . Problem Annika z = 0 Denkert y = 0 P 2 Joint work with M. Janssen n points Introduction Preliminaries Main Results Another x = 0 Approach 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 � x a y b z c F d � � � c < n The set is a k -basis for R = k [ x, y, z ] . Containment Problem Annika Lemma Denkert Joint work Let m ≥ 1 . Then with M. Janssen 1. ( x, y ) m = � x a y b z c F d | c < n, a + b ≥ m � Introduction 2. ( xy, F ) m = � x a y b z c F d | c < n, min( a, b ) + d ≥ m � Preliminaries Main Results 3. I ( m ) = ( x, y ) m ∩ ( xy, F ) m Another Approach = � x a y b z c F d | c < n, a + b ≥ m, min( a, b ) + d ≥ m � Lemma Let r ≥ 1 . Then I r = � x a y b z c F d | c < n, a + b ≥ r, min( a, b )+ d ≥ r, a + b + d ≥ 2 r � . 6 / 15
Recap Containment Let m, r ≥ 1 . Then Problem Annika I ( m ) = � x a y b z c F d | c < n, a + b ≥ m, min( a, b ) + d ≥ m � Denkert Joint work with M. Janssen and Introduction I r = � x a y b z c F d | c < n, a + b ≥ r, min( a, b )+ d ≥ r, a + b + d ≥ 2 r � . Preliminaries Main Results The conditions on ( a, b, c, d ) in the description of I r mirror Another Approach those given in the description of I ( m ) , except for that last condition, a + b + d ≥ 2 r . It turns out that this is the condition that determines containment of I ( m ) in I r . 7 / 15
Containment and the Resurgence Containment Problem Theorem (1) Annika Denkert We have I ( m ) � I r if and only if Joint work with M. Janssen (a) either r > m or Introduction (b) r ≤ m and there exists x a y b F d ∈ I ( m ) such that Preliminaries Main Results a + b + d < 2 r . Another Approach Theorem (2) For m, r ≥ 1 we have I ( m ) � I r if and only if 4 r > 3 m + 1 . In particular, the resurgence ρ ( I ) is 4 3 . 8 / 15
Proof of (1) ⇒ (2) Containment Problem � (a) r > m ⇒ 4 r > 4 m ≥ 3 m + 1 Annika Denkert (b) r ≤ m , some x a y b F d ∈ I ( m ) with a + b + d < 2 r : Joint work with M. Janssen 4 r > 2( a + b + d ) ≥ 3 m as Introduction a + b ≥ m a + d ≥ m b + d ≥ m Preliminaries m odd: Main Results 4 r > 2( a + b + d ) ≥ 3 m ⇔ 4 r > 2( a + b + d ) ≥ 3 m + 1 Another Approach � m even: 4 r > 3 m ⇔ 2 r > 3 m/ 2 ⇔ 2 r ≥ 3 m/ 2 + 1 ⇔ 4 r ≥ 3 m + 2 � ⇔ 4 r > 3 m + 1 9 / 15
Conjectures Containment Conjectures Problem Let R = k [ P N ] , M = ( x 0 , . . . , x N ) the irrelevant ideal, and Annika Denkert J � R a nontrivial ideal. Then for all m, r ∈ N Joint work with M. Janssen ◮ (B. Harbourne, C. Huneke) If J is a radical ideal of a finite set of points, then J ( Nr ) ⊆ M r ( N − 1) J r and Introduction J ( Nr − N +1) ⊆ M ( r − 1)( N − 1) J r . Preliminaries Main Results ◮ (B. Harbourne, C. Huneke) If J is a radical ideal of a finite Another Approach set of points, then J ( r ( m + N − 1)) ⊆ M r � J ( m ) � r . ◮ (B. Harbourne) If J is homogeneous, then J ( Nr − N +1) ⊆ J r . 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 M t I r for any t, r ∈ N 0 . 10 / 15
I ( m ) ⊆ M t I r Containment Problem Let M = ( x, y, z ) . Annika To find out for which ( m, t, r ) we have I ( m ) ⊆ M t I r , we need Denkert Joint work to assume that I ( m ) ⊆ I r . Therefore, we need to assume that with M. Janssen 4 r ≤ 3 m + 1 . Introduction Let r max denote the largest r such that I ( m ) ⊆ I r , i.e. Preliminaries r max = m − ⌈ m − 1 4 ⌉ . Main Results Another Approach We can write M t I r = � x a 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 ) ≥ 2 r, and α + β + nδ ≥ max(0 , t − c ) � . This description is helpful in proving the following theorem. 11 / 15
I ( m ) ⊆ M t I r Containment Theorem Problem Annika Let m ∈ N . Denkert Joint work If r = r max , then with ◮ I ( m ) ⊆ I r but I ( m ) � M I r if m ≡ 0 , 1 mod 4 . M. Janssen Introduction ◮ I ( m ) ⊆ M t I r but I ( m ) � M t +1 I r , where Preliminaries t = min( n, 2 m − 2 r ) if m ≡ 2 , 3 mod 4 . Main Results Another If r < r max , then Approach ◮ I ( m ) ⊆ M ⌈ 3 m 2 ⌉− 2 r I r but I ( m ) � M ⌈ 3 m 2 ⌉− 2 r +1 I r if n = 1 . ◮ I ( m ) ⊆ M t I r but I ( m ) � M t +1 I r , where ◮ t = 2 m − 2 r if r ≤ ⌊ m 2 ⌋ and n ≥ 2 . � � ◮ t = n ⌈ 3 m 2 ⌉ − 2 r if r > ⌊ m 2 ⌋ and n ≥ 2 . 12 / 15
Symbolic Powers as Ordinary Powers Containment Problem Lemma Annika Denkert For any s, t ∈ N , we have Joint work ◮ I (2 st ) = ( I (2 s ) ) t and with M. Janssen ◮ I (2 s + t ) = I (2 s ) I ( t ) . Introduction Preliminaries In fact, we can re-write all symbolic powers of I as follows. Main Results Another Proposition Approach Let m ∈ N . Then I (2) � m ◮ I ( m ) = � 2 if m is even. I (2) � m − 1 ◮ I ( m ) = � I if m is odd. 2 13 / 15
I ( m ) and I r as sums of ideals Containment We can also write I ( m ) and I r as sums of ideals, Problem Annika Denkert m Joint work � I ( m ) = ( xy ) i ( x, y ) max(0 ,m − 2 i ) F m − i with M. Janssen � �� � i =0 = J i Introduction Preliminaries and r Main Results � I r = ( xy ) j ( x, y ) r − j F r − j . Another Approach � �� � j =0 = K j 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 J i ⊆ K j . 14 / 15
Where do we go from here? Containment Problem We are trying to use this or the vector space approach to find Annika Denkert the resurgence in the following case. Joint work with M. Janssen z = 0 Introduction y = 0 P 2 Preliminaries Main Results n points Another Approach x = 0 n points 15 / 15
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