A. V. Geramita and E. Carlini asked that their slides be combined - PDF document

A. V. Geramita and E. Carlini asked that their slides be combined into a single file, since their talks were meant to be taken as parts 1 and 2 of a single presentation.

. Solution to Warings’s Problem for Monomials - I M.V. Catalisano, E. Carlini, A.V. Geramita 1

Waring’s Problem in Number Theory Begins with i ) Lagrange’s observation that every integer is a sum of ≤ 4 squares of integers. ii ) Gauss’ observation that n ≡ 7( mod 8) is not a sum of three squares. Waring asserts (and Hilbert proves) that: there are integers g ( j ) such that every integer is a sum of ≤ g ( j ) j th powers. In particular, Waring asserts that g (3) = 9 etc. That is proved but, unlike Gauss’ observation, only 23 and 239 need nine cubes. Waring’s Second Problem: Find G ( j ), the least positive integers so that every sufficiently large integer is a sum of ≤ G ( j ) j th powers. 2

Are there analogs to Waring’s Problems in S = C [ x 1 , . . . , x n ] = ⊕ ∞ i = o S i ? Yes-1, Lagrange analog. Let F ∈ S 2 , then F = L 2 1 + · · · + L 2 k , k ≤ n. Moreover, almost every F is a sum of n squares of linear forms. Those which require fewer lie on a hypersurface in P ( S 2 ). Yes-2, Hilbert analog. Let t = dim S d . There are linear forms L 1 , . . . , L t such that L d 1 , · · · , L d t are a basis for S d . Yes-3, Waring Cubes analog. Let S = C [ x 1 , x 2 ], P 3 = P ( S 3 ). i ) the points of P 3 of the form [ L 3 ] are the rational normal curve, C , in P 3 . ii ) The points [ F ] not on C , but on its tangent envelope, require 3 cubes. iii ) The general point [ F ] in P 3 , i.e. a point not on the tangent envelope, requires 2 cubes. 3

Definition: Let F ∈ S d , a Waring Decomposition of F is a way to represent F = L d 1 + · · · + L d s such that no shorter such representation exists. In this case we say that the (Waring) rank of F is s . In 1995, J. Alexander and A. Hirschowitz solved the long outstanding prob- lem of finding the Waring rank of a general form in S d for any d and any n . (roughly speaking, it is on the order of dim S d / ( n + 1) ) . However, it is hard to know when one has a general form! and, as we saw, the Waring Rank of a specific form can be larger than the general rank. There is a way to find the rank of any specific form, and this involves the use of Macaulay’s Inverse System. 4

Let F ∈ S d , S = C [ x 1 , . . . , x n ] and let T = C [ y 1 , · · · , y n ]. We make S into a graded T -module by y i ◦ F = ( ∂/∂x i )( F ) and extend linearly. Definition: Given F ∈ S d , then F ⊥ = { ∂ ∈ T | ∂F = 0 } . It is easy to see that F ⊥ is a homogeneous ideal in T . Less obvious is the fact that it is always a Gorenstein Artinian ideal in T . Apolarity Lemma: F ∈ S d and I = F ⊥ ⊂ T . If we can find J ⊂ I where J = ℘ 1 ∩ · · · ∩ ℘ s is the ideal of a set of s distinct points in P n − 1 , then F = L d 1 + · · · + L d s . 5

So, it’s enough to find the smallest set of distinct points in P n − 1 whose defining ideal is in F ⊥ . There have been several attempts to calculate the Waring Rank of specific forms. In particular, Landsberg-Teitler and Schreyer-Ranestad (among others) have attempted to find the Waring rank of monomials (and succeeded for certain monomials). Theorem: (Catalisano, Carlini, G..) Let F = x b 1 1 x b 2 2 · · · x b n n . Then the Waring rank of F is exactly s = ( b 2 + 1)( b 3 + 1) · · · ( b n + 1) . By the Apolarity Lemma, it will be enough to show that i ) F ⊥ contains an ideal of s distinct points; and ii ) F ⊥ does not contain an ideal with fewer than s points. 6

The first part is simple: It comes from the observation that F ⊥ = ( y b 1 +1 , y b 2 +1 , · · · , y b n +1 ) n 1 2 in the first instance, and that F 1 = y b 2 +1 − y b 2 +1 − y b n +1 , · · · , F n − 1 = y b n +1 n 2 1 1 is a regular sequence in T which defines a complete intersection of s distinct points. The more difficult (and interesting) part of the proof is left to Carlini. 7

The solution to the Waring problem for monomials - II E. Carlini Dipartimento di Matematica Politecnico di Torino, Turin, Italy AMS 2011 Fall Central Section Meeting, University of Nebraska-Lincoln, November 2011 1/20 E. Carlini The solution to the Waring problem for monomials - II

Joint work E.Carlini, M.V.Catalisano, A.V.Geramita The solution to the Waring problem for monomials. 4th October 2011, arXiv:1110.0745v1 2/20 E. Carlini The solution to the Waring problem for monomials - II

Apolarity Lemma For a degree d form F ∈ S d one can write s � L d F = i i = 1 if and only if there exists a set of s distinct points X ⊂ P ( S 1 ) such that I X ⊂ F ⊥ . 3/20 E. Carlini The solution to the Waring problem for monomials - II

The case of monomials Consider the monomial M = x b 1 1 · . . . · x b n n where 1 ≤ b 1 ≤ . . . ≤ b n and notice that M ⊥ = ( y b 1 + 1 , . . . , y b n + 1 ) . n 1 Thus we want to study the multiplicity of one dimensional radical ideals I such that I ⊂ ( y b 1 + 1 , . . . , y b n + 1 ) . n 1 4/20 E. Carlini The solution to the Waring problem for monomials - II

Ideal of points in ( y a 1 1 , . . . , y a n n ) So we study monomial ideals generated by powers of the variables. Notice that ( y a 1 1 , . . . , y a n n ) contains the ideal I X = ( y a 2 2 − y a 2 1 , . . . , y a n n − y a n 1 ) and this is the ideal of a set of points X which is a complete intersection consisting of Π n 2 ( a i ) distinct points. Of course we can find larger set of points, but can we find smaller sets? 5/20 E. Carlini The solution to the Waring problem for monomials - II

Main Theorem We proved the following Theorem Let n > 1 and K = ( y a 1 1 , . . . , y a n n ) be an ideal of T with 2 ≤ a 1 ≤ . . . ≤ a n . If I ⊂ K is a one dimensional radical ideal of multiplicity s , then n � s ≥ a i . i = 2 Thus, if X is set of s distinct points such that I X ⊂ K , then s ≥ � n i = 2 a i . 6/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof We work out an example. Let K = ( y 2 1 , y 3 2 , y 4 3 ) and we look for ideal of points I X ⊂ K . Clearly I X = ( y 3 2 − y 3 1 , y 4 3 − y 4 1 ) ⊂ ( y 2 1 , y 3 2 , y 4 3 ) = K and X is a set of 12 distinct points. We want to show that there is no set of less than 3 × 4 distinct points such that I X ⊂ K . 7/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof Radical (i.e. distinct points) is essential Remark Notice that K = ( y 2 1 , y 3 2 , y 4 3 ) ⊃ ( y 2 1 , y 3 2 ) where the latter is a one dimensional not radical ideal of multiplicity 6. Hence the result only holds for sets of distinct points. 8/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof To bound the multiplicity of I X ⊂ K we bound its Hilbert function as � R � HF , t ≤ | X | I X for all t . We also notice that I X + ( y 2 1 ) ⊂ K = ( y 2 1 , y 3 2 , y 4 3 ) and hence � � � R � R HF 1 ) , t ≥ HF K , t . I X + ( y 2 9/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof We now consider to cases depending on whether y 1 is a 0-divisor in R I X . If y 1 is not a zero divisor in R I X . Hence � � � R � � R � R HF 1 ) , t = HF , t − HF , t − 2 I X + ( y 2 I X I X and we get the relation � R � � R � � R � HF , t ≥ HF K , t + HF , t − 2 I X I X using this expression we obtain the desired bound on | X | . 10/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof K = ( y 2 1 , y 3 2 , y 4 3 ) is a complete intersection thus 0 1 2 3 4 5 6 HF ( R / K , · ) = 1 3 5 6 5 3 1 Now we iterate the relation � R � � R � � R � HF , 6 ≥ HF K , 6 + HF , 4 I X I X � R � � R � � R � HF , 6 ≥ 1 + HF K , 4 + HF , 2 I X I X � R � � R � HF , 6 ≥ 1 + 5 + HF , 2 I X I X 11/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof 3 ) and y 1 is not a zero divisor in R As I X ⊂ ( y 2 1 , y 3 2 , y 4 I X , we have � R � HF , 2 = 6 I X and hence � R � HF , 6 ≥ 1 + 5 + 6 = 12 I X which proves | X | ≥ 12. 12/20 E. Carlini The solution to the Waring problem for monomials - II

Idea of the proof If y 1 is a zero divisor in R I X . Consider the ideal I Y = I X : ( y 1 ) and notice that I Y ⊂ K : ( y 1 ) = ( y 1 , y 3 2 , y 4 3 ) . As y 1 is not a 0-divisor in R I Y we can use the same argument of the previous case and we get | X | > | Y | ≥ 3 × 4 . 13/20 E. Carlini The solution to the Waring problem for monomials - II

Consequences The rank of any monomial. Corollary For integers m > 1 and 1 ≤ b 1 ≤ . . . ≤ b m let M be the monomial x b 1 1 · . . . · x b m m then rk ( M ) = � m i = 2 ( b i + 1 ) , i.e. M is the sum of � m i = 2 ( b i + 1 ) power of linear forms and no fewer. 14/20 E. Carlini The solution to the Waring problem for monomials - II

Remark After we posted our paper on the arXiv we received a draft from W. Buczynska, J. Buczynski and Z. Teitler. This draft contains a statement giving an expression for the rank of any monomial coinciding with the one that we found. 15/20 E. Carlini The solution to the Waring problem for monomials - II