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 Mike Janssen (UNL) Containment Problem October 16, 2011 1 / 13

Introduction The Problem Recall: Q: Given an ideal I , how do I ( m ) and I r compare? If I ⊆ R = k [ P N ] is the ideal of points p 1 , p 2 , . . . , p d ∈ P N , then I ( m ) = ∩ i I ( p i ) m . Example: If I defines a complete intersection, I ( m ) = I m . Facts: Given 0 � = I � R = k [ P N ] 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 2 / 13

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: � m / r : I ( m ) �⊆ I r � ρ ( I ) = sup Obviously, m / r > ρ ( I ) implies I ( m ) ⊆ I r . Mike Janssen (UNL) Containment Problem October 16, 2011 3 / 13

Introduction Some facts about ρ ( I ) Theorem (Hochster-Huneke) For I ⊆ k [ P N ] homogeneous, I ( rN ) ⊆ I r . Corollary If I ⊆ k [ P N ] is nontrivial and homogeneous, 1 ≤ ρ ( I ) ≤ N. For ideals 0 � = I � k [ P N ] 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 4 / 13

Introduction Notation Points in P 2 , ideals in R = k [ P 2 ] = k [ x , y , z ]: y = 0 x = 0 p 0 z = 0 p 1 p 2 p 3 · · · p n n ≥ 3 I ( p 0 ) = ( x , y ) and I ( p 1 + · · · + p n ) = ( z , F ), F ∈ k [ x , y ], deg F = n I = I ( p 0 + p 1 + · · · + p n ) = ( 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 October 16, 2011 5 / 13

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 = span k � x β y γ z δ � 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 . Mike Janssen (UNL) Containment Problem October 16, 2011 6 / 13

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. Mike Janssen (UNL) Containment Problem October 16, 2011 7 / 13

Preliminaries An Example Claim: I (7) �⊆ 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 8 / 13

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 ≤ rn 2 + rn − n − 2 , or n 2 n 2 r − n m ≥ n and m ≤ n 2 − n + 1 . Theorem (Asymptotic Solution) For the ideal I of n ≥ 3 collinear points and one point off the line, n 2 ρ ( I ) = n 2 − n + 1 . Mike Janssen (UNL) Containment Problem October 16, 2011 9 / 13

Main Results Computing ρ ( I ) from the complete solution Assume m ≥ n n 2 r − n Recall I ( m ) �⊆ I r ⇔ m ≤ n 2 − n + 1. n 2 − n + 1 > n 2 − n / r n 2 n 2 r − n Then m n 2 − n + 1 ⇒ I ( m ) ⊆ I r . r > n 2 − n + 1 ⇒ m > n 2 Thus, ρ ( I ) ≤ n 2 − n + 1. Conversely, take m = tn 2 − 1 and r = t ( n 2 − n + 1).Then n 2 r − n m ≤ n 2 − n + 1, but tn 2 − 1 n 2 n 2 m lim r = lim t ( n 2 − n + 1) = n 2 − n + 1 , so ρ ( I ) ≥ n 2 − n + 1 . t →∞ t →∞ Mike Janssen (UNL) Containment Problem October 16, 2011 10 / 13

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. Mike Janssen (UNL) Containment Problem October 16, 2011 11 / 13

Applications Two conjectures Harbourne and Huneke conjectured: Conjecture I (2 r ) ⊆ M r I r , where M is the ideal generated by the variables. Conjecture I (2 r − 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. Mike Janssen (UNL) Containment Problem October 16, 2011 12 / 13

Applications Thank you! Mike Janssen (UNL) Containment Problem October 16, 2011 13 / 13