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©Alessandro Oneto, Stockholm 2014e-mail: [email protected]
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Abstract
This thesis is divided into two chapters. First, we want to study particularclasses of power ideals, with particular attention to their relation with theFröberg conjecture on the Hilbert series of generic ideals. In the second part,we study a generalization (introduced in Fröberg, Ottaviani, and Shapiro [25])of the classical Waring problem for polynomials about writing homogeneouspolynomials as sums of powers. We see also how the theories of fat points andsecant varieties of Veronese varieties play a crucial role in the relation betweenthose chapters and in providing tools to �nd an answer to our questions.
The main results are the computation of the Hilbert series of particularclasses of power ideals, which in particular give us a proof of the Fröberg con-jecture for generic ideals generated by eight homogeneous polynomials of thesame degree in four variables, and the solution of the generalized Waring prob-lem in the case of sums of squares in three and four variables. We also beginthe study of the generalized Waring problem for monomials.
Sammanfattning
Denna avhandling har två kapitel. I det första studerar vi en speciell klass avpower ideals (ideal genererade av potenser av linjärformer). Speciellt studerarvi sambandet med Fröbergs förmodan om Hilbertserier för generiska ideal. Idet andra studerar vi en generalisering introducerad i [25] av det klassiskaWaringproblemet för polynom som behandlar problemet att skriva homogenapolynom som summor av potenser. Vi visar också hur teorin om feta punkter
och sekantvarieteter av Veronesevarieteter spelar en väsentlig roll i relationenmellan de två kapitlen och i möjligheten att få svar på våra frågor.
Huvudresultaten är beräkningen av Hilbertserier av vissa klasser av powerideals, vilket speciellt ger oss ett bevis för Fröbergs förmodan för generiskaideal genererade av åtta former av samma grad i fyra variabler, och lösningentill det generaliserade Waringproblemet för summor av kvadrater i tre och fyravariabler. Vi behandlar också det generaliserade Waringproblemet för monom.
Acknowledgements
Firstly, I would like to thank my supervisors Prof. Boris Shapiro and Prof. RalfFröberg for the instructive and stimulating mathematical conversations and fortheir accurate attention on my work.
I would like also to acknowledge Prof. Enrico Carlini and Prof. Jörgen Back-elin for the possibility of working together. I want to express my gratitude alsoto Prof. Giorgio Ottaviani, Prof. Maria Virginia Catalisano and Prof. AnthonyGeramita for their comments during the writing of this thesis and the usefuldiscussions about possible future projects related to it.
I have to deeply thank my parents, Teresa and Sera�no, and my girlfriendChiara for their support since I made the decision of moving to Stockholm.Without them, these years would have been much more di�cult and eventhe results included in this thesis would have been much more complicatedto achieve.
Finally, I want to thank all my friends: the new ones, sharing with me thisnew adventure with their helpful words and smiles, and the old ones that I feelso close to me even if hundreds kilometers far away.
To Tony Geramita,who made me start this long path
List of Papers
The following papers are included in this thesis.
PAPER I: On a class of power idealsJörgen Backelin, Alessandro Oneto,arXiv preprint: 1403.4793.
PAPER II: Monomials as sum of kth-powersEnrico Carlini, Alessandro Oneto,arXiv preprint: 1305.4553.Accepted in Communication in Algebra.DOI: 10.1080/00927872.2013.842247
Contents
Abstract v
Sammanfattning vii
Acknowledgements ix
List of Papers xiii
Preface xvii
1 Power ideals and the Fröberg conjecture 19
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.1.1 Basic de�nitions . . . . . . . . . . . . . . . . . . . . . . . . . . 191.1.2 Fröberg conjecture . . . . . . . . . . . . . . . . . . . . . . . . . 211.1.3 Fat points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.1.4 Macaulay duality . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.2 8 generic fat points in P3 (with Ralf Fröberg) . . . . . . . . . . . . . . 291.3 PAPER I : A class of power ideals (with Jörgen Backelin) . . . . . . . 38
1.3.1 Multicycle gradation . . . . . . . . . . . . . . . . . . . . . . . . 381.3.2 Hilbert function of the power ideal In,k,d . . . . . . . . . . . . . 421.3.3 Hilbert function of ξ-points in Pn . . . . . . . . . . . . . . . . . 52
2 The Waring problem for polynomials 61
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 612.1.1 Historical background . . . . . . . . . . . . . . . . . . . . . . . 612.1.2 The geometry of the problem . . . . . . . . . . . . . . . . . . . 63
2.2 Sum of squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.2.1 A generalized Waring problem . . . . . . . . . . . . . . . . . . 682.2.2 Sum of squares in three variables. . . . . . . . . . . . . . . . . . 702.2.3 Sum of squares in four variables. . . . . . . . . . . . . . . . . . 72
2.3 A geometric approach . . . . . . . . . . . . . . . . . . . . . . . . . . . 752.4 PAPER II : Monomials as sum of kth-powers (with Enrico Carlini) . . 78
2.4.1 Basic facts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 782.4.2 Results on the kth-rank for monomials . . . . . . . . . . . . . . 802.4.3 Final remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
References lxxxv
Preface
The aim of this thesis is to resume part of my research work competed duringmy �rst period at Stockholm University. The thesis aims to be self-contained,trying to give the history and the background required to understand theproblems studied. However, when some basic notion is omitted or partiallydescribed, references are given.
The thesis is divided into two chapters.In Chapter 1, we study power ideals, with particular attention to the con-
nections with the Fröberg conjecture about Hilbert series of generic ideals.
In a polynomial ring, ideals generated by powers of linear forms are calledpower ideals. In the last decades, they have been largely studied because of theirconnection with many di�erent �elds of mathematics. We focus on the relationwith ideals of fat points. In particular, via Macaulay duality, the Hilbert seriesof power ideals is strictly related to the Hilbert series of ideals of fat points.We also see how the study of Hilbert series of power ideals and fat points cangive us an answer for the Fröberg conjecture in some particular cases.
In Section 1.1, we introduce the basic de�nitions and some historical back-ground on the Fröberg conjecture, fat points and Macaulay duality.
In Section 1.2, we consider the polynomial ring in four variables and wefocus on the power ideals generated by dth-powers of linear forms which arethe sum of an odd number of variables. By studying the associated scheme offat points, we prove that such power ideals are su�ciently generic and theyhave the Hilbert series described by the Fröberg conjecture. This is enough toprove that the conjecture holds for 8 generators of degree d in four variables.
In Section 1.3, we focus on a particular family of power ideals. For �xedpositive integers k, d with k ≥ 2, let ξ be a primitive kth-root of unity in C.In the polynomial ring C[x0, . . . , xn], we consider the ideals In,k,d generatedby the (k − 1)dth-powers of the kn powers of linear forms (x0 + ξg1x1 + . . . +ξgnxn)(k−1)d with 0 ≤ gi ≤ k− 1, for all i = 1, . . . , n. By using a Zn+1
k -gradingon the polynomial ring, we study the structure of such ideals and, for thek = 2 case, it gives us a numerical algorithm to compute the Hilbert functionof the corresponding quotient rings. We conjecture the extension of such analgorithm for k > 2. On the other hand, we compute the Hilbert function ofthe same quotient rings by computing the Hilbert function of the corresponding0-dimensional schemes with support on the kn points [1 : ξg1 : . . . : ξgn ] ∈ Pn.That this result agrees with the conjectured algorithm for k > 2 is supported
by several computer experiments.
In Chapter 2, we look at the Waring problem for polynomials.
Given an homogeneous polynomial of degree k, what is the minimal number of
linear forms needed to write it as sum of their kth-powers?
After an introduction in Section 2.1, on the history behind this classical ques-tion and the connections with other problems involving fat points or secantvarieties, we look at the work started in Fröberg, Ottaviani, and Shapiro [25]by investigating a generalization of the Waring problem for polynomials.
Given an homogeneous polynomial of degree kd, what is the minimal number
of forms of degree d needed to write it as sum of their kth-powers?
Their main result in Fröberg, Ottaviani, and Shapiro [25] is an upper boundfor the number of summands required by the generic polynomial. Moreover, thisbound doesn't depend on the parameter d and, asymptotically with respect tod, is sharp. Our project is to continue the work started in that paper.
In Section 2.2, we consider sum of squares, namely we �x k = 2. In the caseof three and four variables, we complete the work of Fröberg, Ottaviani, andShapiro [25] by computing the correct answer of the problem for low degrees dbefore becoming equal to the bound 2n.
In Section 2.3, we describe the geometric picture behind this generalizedWaring problem. This section wants to look at a possible direction for a futureproject.
In Section 2.4, we consider the particular case of monomials. The classicalWaring problem has been solved in the case of monomials in Carlini, Catalisano,and Geramita [11] and we try to use their result for our generalization. We givethe complete answer for monomials in three variables.
CHAPTER 1
Power ideals and the Fröberg conjecture
Section 1.1
Introduction
1.1.1 Basic de�nitions
Let S = C[x0, . . . , xn] be the polynomial ring in n + 1 variables with complexcoe�cients. Usually, we consider S as equipped with the standard gradation,i.e. we write S =
⊕i∈N Si; where Si denotes the C-vector space of homogeneous
polynomials of degree i in S.
De�nition 1.1. A homogeneous ideal I ⊂ S is called a power ideal if I isgenerated by a collection of powers of linear forms which generate S1.
This class of ideal has recently received considerable attention in the mathe-matical literature thanks to the connection with many di�erent areas of com-mutative algebra, algebraic geometry or combinatorics. For a complete andclear survey about these connections look at Ardila and Postnikov [6].
Given a power ideal I ⊂ S, since it is homogeneous, we have that thequotient algebra R = S/I is also graded, i.e. R =
⊕i∈NRi and it makes sense
to de�ne the Hilbert function and the Hilbert series of the power ideal I andits quotient ring R.
De�nition 1.2. LetM =⊕
i∈N be a graded S-module. We de�ne its Hilbertfunction to be
HF(M ; i) := dimCMi, for all i ∈ N;
from this function, we can even associate to the module M its Hilbert serieswhich is the power series de�ned as
HSM (t) :=∑i∈N
HF(M ; i)ti.
19
Example 1.3. Let I = (x30, x21, x
22) ⊂ C[x0, x1, x2]. Then, the quotient ring
R = C[x0, x1, x2]/I is generated as a C-vector space by
1,x0, x1, x2,x20, x0x1, x0x2, x1x2x20x1, x20x2 x0x1x2,
=⇒
HF(R; 0) = 1;HF(R; 1) = 3;HF(R; 2) = 4;HF(R; 3) = 3;HF(R; i) = 0, for all i ≥ 4.
and the Hilbert series is HSR(t) = 1 + 3t+ 4t2 + 3t3.
Before going further, we want to recall a few basic algebraic notions 1. A gradingover the polynomial ring can be de�ned over any monoid; hence, the standardgrading on our polynomial ring S can be thought as a grading over Z. In thisway, it makes sense to de�ne the shifting of an S-module.
De�nition 1.4. Let M =⊕
i∈ZMi be a graded S-module. For any integer j,we de�ne the shifted moduleM(j) as the same S-moduleM with the gradedstructure given by
[M(j)]i := Mi−j , for all i ∈ Z.
In the category of graded S-modules, we want that the maps are morphisms ofS-modules which preserve the grading. In other words, given graded S-modulesM and N and a morphism of S-modules f : M → N , we say that f is gradedof degree j if, for any i ∈ Z,
f(Mi) ⊂ Ni+j .
From these de�nitions, it clear that, given a graded map f : M → N of degreej, we can easily consider it as a degree 0 map by shifting the module M by −j.Degree 0 maps are very useful in computing Hilbert functions of S-modules.Indeed, given an exact sequence of graded S-modules with degree 0 gradedmaps
0 // M // N // P // 0we have that,
HF(N ; i) = HF(M ; i) + HF(P ; i), for all i ∈ Z.
Example 1.5. (Complete intersections) Given an S-module M , we saythat a non-zero element f ∈ S is M-regular if f is a non-zero divisor on M(for short NZD), i.e. if fg = 0 for g ∈ M then g = 0. In general, a sequence(f1, . . . , fg) of elements in S is called an M-regular sequence, or simply anM -sequence, if the following conditions are satis�ed:
1. fi is M/〈f1, . . . , fi−1〉M -regular element for all i = 1, . . . , g;
2. M/〈f1, . . . , fg〉M is non zero.
1See Bruns and Herzog [10] for an extended explanation.
20
A quotient ring R = S/I is called a complete intersection if I is a homoge-neous ideal generated by an S-sequence (f1, . . . , fg). Under these assumptions,the computation of the Hilbert series of R is an easy exercise. We proceed byinduction on the number of generators. Assume g = 1 and deg f1 = d1; hence,we get the following exact sequence with graded maps of degree 0.
0 // S(−d1)·f1 // S // S/I // 0.
Hence, we get that HF(S/I; i) = HF(S; i)− HF(S; i+ d1), for all i ∈ Z. Thus,looking at their Hilbert series, we get
HSS/I(t) = (1− td1) HSS(t) = (1− td1)∑i∈N
(n+ i
n
)ti =
1− td1(1− t)n+1
.
Now, let I = (f1, . . . , fg) with g > 1, then, by induction, from the exactsequence of degree 0 maps
0 // S/(f1, . . . , fg−1)(−dg)·fg
// S/(f1, . . . , fg−1) // S/I // 0.
Hence, we get, similarly as in the g = 1 case,
HSS/I(t) = (1− tdg ) HSS/(f1,...,fg−1)(t) =
∏gi=1(1− tdi)(1− t)n+1
. (1.1)
1.1.2 Fröberg conjecture
In 1985, Ralf Fröberg was studying the Hilbert function of generic ideals.
Any homogeneous polynomial of degree d in S = C[x0, . . . , xn] is a linearcombination of the monomials of degree d in n + 1 variables; hence, it can beseen as a point of an a�ne space AN of dimension N =
(n+dn
). We say that
a property P holds for generic forms if there exists a Zariski (dense) opensubset U in AN such that P holds for all forms in U .1 Similarly, if we considera homogeneous ideal I with numerical character (n, d1, . . . , dg), namely I isgenerated by homogeneous polynomials in n + 1 variables f1, . . . , fg of degreedi for i = 1, . . . , g, respectively; then, I can be seen as a point in the productof a�ne spaces AN1 × . . . × ANg with Ni =
(n+din
). We say that a property P
holds for a generic ideal if there exists a Zariski open (dense) subset U whereP holds for any ideal in U .In Fröberg and Löfwall [24], the authors proved that, �xing a numerical char-acter, there exists an open subset U of the a�ne space AN1 × . . .×ANg wherethe smallest Hilbert series is attained. For example, considering ideals with nu-merical character with g ≤ n + 1, we have that a generic ideal is a completeintersection and we have computed its Hilbert series in Example 1.5. There is anatural guess for the Hilbert series of generic ideals with some given numericalcharacter, by trying to extend the formula (1.1). In Fröberg [21], the author
1The Zariski topology on a�ne or projective spaces is the basic topology compatible with
Algebraic Geometry. Closed subspaces are de�ned as zero loci of polynomials. They are calledvarieties and their complements, the open subspaces, are always dense. See Harris [29] for basicnotions of Algebraic Geometry.
21
proved that, given any ideal I with numerical character (n, d1, . . . , dg), we havethat
HSS/I(t) �Lex
⌈∏gi=1(1− tdi
(1− t)n+1
⌉; (1.2)
where the inequality has to be understood in the lexicographic sense and where⌈∑i ait
i⌉is de�ned as the power series
∑i bit
i with bi := ai, if aj > 0 for allj ≤ i and bi := 0 otherwise. In the same paper, Ralf Fröberg conjectured that,for generic ideals, the inequality (1.2) is an equality.
Conjecture 1 (Fröberg Conjecture, 1985). For a generic ideal I of nu-
merical character (n, d1, . . . , dg),
HSS/I(t) =
⌈∏gi=1(1− tdi
(1− t)n+1
⌉. (1.3)
Even if the latter conjecture attracted the attention of many algebraist to tryto prove it, up to now, only a very few cases are proven. As we have seen, theg ≤ n + 1 follows from the fact that generic ideals are complete intersections.Richard Stanley proved the g = n+ 2 case, see Fröberg [21, Example 2, p.127];moreover, the conjecture was proven to be true even for the n = 1 case, seeFröberg [21], and the n = 2 case, see Anick [5].
From the results in Fröberg [21] and Fröberg and Löfwall [24] explainedabove, for any �xed numerical character (n, d1, . . . , dg), it is enough to �ndan ideal with the conjectured Hilbert series in order to prove Conjecture 1.Geometrically speaking, we simply have to show that the open subset wherethe minimal Hilbert series is attained is not empty. We say that an ideal withthe conjectured Hilbert series is Hilbert generic.
Since the main object of study of this thesis are power ideals, there is a verynatural question arising from this introduction to Fröberg conjecture.
Question 1. Are powers of linear forms Hilbert generic?
This question is very interesting. The connections between power ideals andother algebraic objects can give us many tools to attack this problem or, theother way around, we can get results in di�erent �elds by answering this ques-tion about power ideals. Moreover, if we are able to give a positive answer forsome numerical character (n, d1, . . . , dg), we get a proof for the Fröberg conjec-ture. In this case, unfortunately, the answer to Question 1 is far from being truein general. In Fröberg and Hollman [22], the authors used computer calcula-tions, with the help of the �rst version of the software Macaulay, see Bayer andStillman [7], in order to support the Conjecture 1. The idea was basically tocomputer the Hilbert series of a random ideal with a given numerical characterand check wheter it is equal to the conjectured series. Unfortunately, the timeand the memory of a computer are not unlimited and then they have been ableto check the Conjecture only for n+ 2 ≤ g ≤
(d+n−1
d
), with di = 2, n ≤ 11 and
di = 3, n ≤ 8. They also checked that, in those range of cases, power ideals areHilbert generic with a list of exceptions.
22
1.1.3 Fat points
Consider the n-dimensional projective space Pn, i.e. the projective space as-sociated to the C-vector space S1 of our polynomial ring S = C[x0, . . . , xn].1
For any homogeneous ideal of S, we de�ne a variety in Pn as the zero locus ofthe polynomials contained in the radical of the ideal. For example, prime idealsare associated to irreducible varieties. The height of the ideal is the algebraicnotion for the dimension of the variety and, since the unique homogeneousmaximal ideal corresponds, in the projective space, to the empty set, we de�nethe dimension of the variety as one less than the height of the correspondingideal.
A point P ∈ Pn, is the variety associated to a prime ideal ℘ inside ourpolynomial ring S with height n. The ideal ℘ is de�ned as the ideal of homoge-neous polynomials of S vanishing at P . Geometrically speaking, ℘ is the idealof hypersurfaces, i.e. zero loci of principal ideals, which pass through P . TheHilbert function of ℘ in some degree d corresponds simply to the dimension ofthe vector space of degree d hypersurfaces passing through P . More generally,we can compute the Hilbert function of a �nite set of distinct points.
Example 1.6. [Geramita and Orecchia [27]]Hilbert function of s distinct points. Let X = P1 + . . .+Ps be the set of sdistinct points in Pn. As a variety, it is associated to the ideal I = ℘1 ∩ . . .∩℘swhere ℘i is the prime ideal associated to the point Pi. We want to computethe Hilbert function of the ideal I in degree d.
This becomes easily a linear algebra problem. Consider {M1, . . . ,MN}, withN =
(n+dd
), the degree d monomials in n+1 variables, i.e. a basis for the vector
space Sd. A homogeneous polynomial of degree d is then given by a linearcombination F =
∑Nj=1 cjMj . Now, we want to impose the vanishing at each
points P1, . . . , Ps, i.e. we want to solve the linear system associated to thematrix
Md =
M1(P1) M2(P1) . . . MN (P1)M1(P2) M2(P2) . . . MN (P2)
......
. . ....
M1(Ps) M2(Ps) . . . MN (Ps)
.Thus, we have that
HF(S/I; d) = dimSd − dim (solution ofMd) = rankMd.
In general, we can �nd a set of points P1, . . . , Ps such that all the matricesMd
have the maximal rank. Since having the maximal rank is an open condition,we can conclude that a generic set of points X = P1 + . . .+Ps in Pn associatedto the ideal I = ℘1 ∩ . . . ∩ ℘s has the maximal Hilbert function, i.e.
HF(S/I; d) = min
{s,
(n+ d
n
)}.
1Given a C-vector space V of dimension n+ 1, we de�ne the associated projective space as
the space of lines in V ; more precisely, as the quotient of V r {0} by the relation where, forany v, w ∈ V , v ∼ w if and only if v = λw for any λ ∈ C r {0}.
23
From our geometric intuition, we already have the naïve concept of multiple
points. Given a line intersecting a plane conic, let's imagine that we move theline up into tangent position as in Figure 1.1. Since high school, we are usedto saying that the tangent meet the conic twice, but we can describe what thatmeans more rigorously by using the notion of scheme. The di�erence with theconcept of variety is that we don't want to consider only the zero locus ofradical ideals.
Figure 1.1: The naïve idea of a multiple point: the intersection of a tangentand a plane conic.
Take an a�ne chart A2{x,y} of our projective plane P2 where our conic has
equation y = x2 and look at the tangent at the origin, with equation y = 0.Then, we have that the intersection is described by the ideal I = (x2 − y, y) =(x2, y). As a variety, it is still the simple point O = (0, 0) associated to itsradical ideal (x, y), but as a scheme, it is the point O with an extra-directionor another point in�nitesimally close: the hypersurfaces contained in I are thehypersurfaces which pass through the point O and their �rst derivative alongthe x-axis vanishes. The simple point O is called the support of the schemeassociated to the ideal I. 1
Following such intuitive description, a fat point of multiplicity 2 can bethought as a point with attached all possible directions or similarly with npoints ini�nitesimally close.
De�nition 1.7. A fat point of multiplicity d in Pn is a 0-dimensionalscheme associated to a dth-power ℘d of a prime ideal ℘. It is denoted by dPand the simple point P de�ned by the ideal ℘ is the support of the fat point.In general, we de�ne a scheme of fat points as X = d1P1 + . . . + dsPs inPn, where P1, . . . , Ps are distinct points, with corresponding ideals ℘1, . . . , ℘s,
1For a precise exposition of the basic notions of the theory of schemes supported with en-
lightening examples, see Eisenbud and Harris [19].
24
respectively, and d1, . . . , ds are positive integers; in other words, X is the 0-dimensional scheme associated to the ideal ℘d11 ∩ . . . ∩ ℘dss .
A homogeneous polynomial F is contained in the ideal ℘d if all the partialderivatives of order ≤ d − 1 of F vanish at the point P . Thus, if we look atthe Hilbert function of ℘d in degree i, we have that, if i < d then we have that[℘d]i = 0. Otherwise, for any i ≥ d, we have to impose
(n+d−1n
)independent
equations on the space of hypersurfaces of degree i in Pn and then we have thatdim[℘d]i = dimSi −
(n+d−1n
). In conclusion, we have that
HF(S/℘d; i) = min
{(n+ i
n
),
(n+ d− 1
n
)}.
Remark 1.8. As our naïve intuition of fat point would suggest, at least withrespect to the Hilbert function, a fat point of multiplicity d in Pn behaves as(n+d−1n
)points collapsed together.
Now, we consider a scheme of fat points X = d1P1 + . . . + dsPs. The corre-sponding ideal ℘d11 ∩ . . . ∩ ℘dss is Cohen-Macaulay of dimension 1 and then theHilbert function of the quotient ring is weakly increasing and, moreover, since∑si=1
(n+di−1
n
)is the maximal number of conditions imposed by X, we also
have that the Hilbert function is eventually equal to∑si=1
(n+di−1
n
). 1 Because
of those properties, one could expect that, at least with respect to the Hilbertfunction, the scheme of fat points X behave like a scheme of
∑si=1
(n+di−1
n
)dis-
tinct points. This would be equivalent to saying that, in each degree j, all theconditions given by the derivatives of order ≤ di at each point Pi are necessaryto determine the hypersurfaces of degree j with the singularities prescribed bythe scheme X. If so, we say that the scheme X impose independent condi-tions on the hypersurfaces of degree j. Unfortunately, it is not hard to �ndsome counter example where the Hilbert function of the scheme X di�ers froma Hilbert function of distinct points.
Example 1.9. Consider the scheme of two distinct double points 2P1 + 2P2
in general position in the projective plane P2. We want to compute the Hilbertfunction of this scheme in degree 2. The space of conics in P2 has (a�ne)dimension 6 and each point imposes exactly 3 condition to the conics. Assumingthat all the conditions are independents, like 6 distinct generic points, we expectto have no conic with two distinct singular points and then the Hilbert functionof the quotient in degree 2 should be equal to 6. However, we have that thedouble line passing through the two points give a conic with two singular points.By Bezout's Theorem, we also have that the double line is also the unique conicand then HF(S/℘2
1 ∩ ℘22; 2) = 5.
Example 1.10. Consider the scheme of �ve double points X = 2P1 + . . .+2P5
in general position in P2. We want to look at its Hilbert function in degree 4.The space of quartics in P2 has (a�ne) dimension
(2+42
)= 15 and each double
point imposes 3 conditions on the space of quartics. Assuming that X behaves
1See Bruns and Herzog [10] for a background on Cohen-Macaulay rings and their properties.
25
as 15 distinct generic points, we expect to have no quartics with �ve singularpoints and the Hilbert function of the quotient in degree 4 should be equal to15. However, there exists a unique conic passing through �ve points and thedouble conic is a quartic inside our ideal ℘2
1 ∩ . . .∩℘25. By Bezout's theorem, it
is also unique and we have that HF(S/℘21 ∩ . . . ∩ ℘2
5; 4) = 14.
With these two examples, we can see that it is very easy to produce casesin which schemes of fat points in general position do not behave as we wouldexpect; hence, the study of their Hilbert functions is far from being obvious andit is still largely unknown. In Geramita and Orecchia [27], the case of simplepoints has been solved, as we have explained in Example 1.6. The case of doublepoints becomes much more complicated, as we have seen in Example 1.9 and1.10, but it has been solved as well.
In 1995, James Alexander and André Hirschowitz, after a series of bril-liant articles, see [2; 3], gave the solution of the problem regarding the Hilbertfunction of double points in general position, see [4].
Theorem 1.11 (Alexander-Hirschowitz Theorem). Let P1, . . . , Ps be
distinct points of Pn in general position, with de�ning ideals ℘1, . . . , ℘s, respec-tively. Then, the scheme of double points X = 2P1 + . . . + 2Ps de�ned by the
ideal ℘21 ∩ . . . ∩ ℘2
s has the expected Hilbert function, i.e.
HF(S/℘21 ∩ . . . ∩ ℘2
s; i) = min
{s(n+ 1),
(n+ i
n
)},
except in the following cases,
(1) i = 2; 2 ≤ s ≤ n;
(2) n = 2; i = 4; s = 5;
(3) n = 3; i = 4; s = 9;
(4) n = 4; i = 4; s = 14;
(5) n = 4; i = 3; s = 7.
Remark 1.12. The cases (2) and (3) corresponds to Examples 1.9 and 1.10.
There is a long bibliography regarding the study of Hilbert function of fatpoints, with enlightening papers investigating particular con�gurations of pointsor some particular cases; however, up to now, simple and double points, arethe only cases completely understood.
1.1.4 Macaulay duality
As we said, we want to focus on the relation between fat points and power ideals.Such relation is given by the Macaulay duality, or Apolarity Lemma, discoveredin Emsalem and Iarrobino [20]. For this part, we refer also to Geramita [26].
26
We consider two di�erent polynomial rings, S = C[x0, . . . , xn] and S =C[X0, . . . , Xn]. We de�ne an action of S on S, where monomials act as partialderivatives 1.
De�nition 1.13 (Apolarity action). For any h ∈ S and f ∈ S, we de�nethe apolarity action of S over S as
h ◦ f := h
(∂
∂X0, . . . ,
∂
∂Xn
)· f(X0, . . . , Xn);
respecting the usual properties of di�erential operators
Example 1.14. Let F = x20 + x1x2 ∈ C[x0, x1, x2] and G = X40 + 2X3
0X2 +X2
1X22 ∈ C[X0, X1, X2]. Then
F ◦G =∂2G
∂X20
+∂2G
∂X1∂X2= 12X2
0 + 12X0X2 + 4X1X2.
De�nition 1.15 (Inverse system). Let I be a graded ring of R, then theinverse system I−1 in S is the annihilator of I, i.e. the S-module
I−1 := {G ∈ S | F ◦G = 0, for all F ∈ I}.
Remark 1.16. The inverse system of a graded ideal is a graded S-module,but, in general, not an ideal since it is not closed under multiplication. Forexample, x2 ◦X = 0, but x2 ◦X2 = 2.
The apolarity action induces a non-degenerate bilinear pairing Si × Si → C;hence, for any graded ideal I in S, we can consider the C-vector space Ii and wecan construct its orthogonal space I⊥i . By de�nition, we have that [I−1]i ⊂ I⊥i ,but actually the equality holds.
Proposition 1.17. Given I a homogeneous ideal in S, for any i ∈ N,
[I−1]i = I⊥i .
Proof. Let F ∈ I⊥i . We only have to prove that H ◦ F = 0 for all H ∈ I.If degH > i, H ◦ F = 0 because the degree of H is bigger than the degree
of F . Assume degH < i and let xα := xα00 · . . . · xαn
n be a monomial of degreei − degH. Hence, xαH ◦ F = 0. By basic properties of di�erentials, we havethat xα ◦ (H ◦F ) = 0 for any monomial of degree i−degF . Since the apolarityaction gives a non-singular pairing, we have that H ◦ F = 0.
Consequently, we have that the study of the inverse system of a homogeneousideal can give informations about the Hilbert function of the ideal. Indeed, forany i ∈ N,
dim[I−1]i = dim I⊥i = dimSi − dim Ii = HF(S/I; i). (1.4)
1In arbitrary characteristic, we should be more careful and consider S as a divided power
ring. Since we are working over C we do not discuss this aspect, but the interested reader can �nda good exposition in Iarrobino and Kanev [31] or in the original paper Emsalem and Iarrobino[20].
27
The idea of Jacques Emsalem and Anthony Iarrobino has been to study theinverse system of schemes of fat points. We give an idea about what is behindtheir result by explaining the case of one point.
Let P0 = [1 : 0 : . . . : 0] ∈ Pn and ℘0 = (x1, . . . , xn). We consider the fatpoint dP0 and we want to compute the inverse system of the ideal ℘d0. Sincewe are in the monomial case, it is enough to understand which monomials arenot inside the ideal ℘d0. Clearly, since ℘
d0 is generated in degree d, we have that
[(℘d0)−1]i = Si for all i < d. For i ≥ d, we have that a monomial is not containedin ℘d0 if and only if it can be divided by xj0 for some j ≥ i− d+ 1, i.e.
[(℘d0)−1]i = Xi−d+10 Sd−1, for all i ≥ d.
In general, given any point P = [p0 : . . . : pn] ∈ Pn and its de�ning ideal ℘,with no loss of generality, we can assume p0 6= 0. After changing coordinatesin S, we can consider new variables Y 's and reduce our computations to thecase of a fat point with support at P0 and use the previous result. Now, it iseasy to see that, by using the inverse change of coordinates, the variable Y0 ismapped into L; hence,
[(℘d)−1]i = Li−d+1Sd−1, for all i ≥ d,
where L = p0X0 + . . .+ pnXn.
Theorem 1.18 (Apolarity Lemma). Let P1, . . . , Ps be distinct points in Pn
with coordinates Pi =[p(i)0 : . . . : p
(i)n
]. Let Li = p
(i)0 X0 + . . .+ p
(i)n Xn, for any
i = 1, . . . , s. Given positive integers d1, . . . , ds, we consider the scheme of fat
points X = d1P1 + . . .+ dsPs de�ned by the ideal I = ℘d1 ∩ . . . ∩ ℘ds, where the
℘i's are the de�ning ideal of the Pi's, respectively. Then,
[I−1]i =
{Si for all i ≤ max{di − 1};Li−d1+11 Sd1−1 + . . .+ Li−ds+1
s Sds−1 for all i ≥ max{di}.
Remark 1.19. The proof follows from our computations of the inverse sys-tem of one fat point and a basic property of inverse system, i.e. given twohomogeneous ideals I and J ,
(I ∩ J)−1 = I−1 + J−1.
The Apolarity Lemma gives a very close relation between fat points and powerideals. Indeed, recalling equation (1.4), we can rephrase it as an equality be-tween Hilbert functions. Consequently, studying the Hilbert function of fatpoints can give us a great deal of information in terms of power ideals and,if we are lucky, give a proof of the Fröberg conjecture. Viceversa, in the caseswhere the Fröberg conjecture is proven, we can use such results to get theHilbert function of the corresponding schemes of fat points.
Corollary 1.20. Let X = d1P1 + . . . + dsPs be a schemes of fat points in Pnwith de�ning ideal I = ℘d11 +. . .+℘dss . Let L1, . . . , Ls be the linear forms de�ned
as in Theorem 1.18. Then,
HF(S/I; i) = HF(
(Li−d1+11 , . . . , Li−ds+1
s ); i), for all i ∈ N.
28
Section 1.2
8 generic fat points in P3(joint work with R. Fröberg 1)
In this section, we want to show an example in which, studying the Hilbertfunction of fat points, we can give a proof of the Fröberg conjecture in somespecial case. In this section, we consider only the uniform cases, i.e. powerideals of numerical character (n, d1, . . . , dg) with di = d, for all i = 1, . . . , g. Wedenote such numerical characters with (n, dg).
First, we look at some numerical property when the number of points is at least2n. In those cases, we have an equivalence between the property of imposingindependent conditions on the hypersurfaces of any degree for the scheme of fatpoints and the property of being Hilbert generic for the corresponding powerideal.
Lemma 1.21. Let d, g and n be positive integers. Assuming g ≥ 2n, then the
power series⌈(1− td)g/(1− t)n+1
⌉is a polynomial of degree ≤ 2d− 2.
Proof. The coe�cient of t2d−1 in (1−zd)g/(1−z)n+1 is equal to the coe�cientof (1 − gtd)/(1 − t)n+1, which is equal to
(2d−1+n
n
)− g(d−1+nn
). Since in our
numerical assumptions 2d+ h < 2(d+ h), for all h ≥ 0, it follows that(2d− 1 + n
n
)− g(d− 1 + n
n
)≤(
2d− 1 + n
n
)− 2n
(d− 1 + n
n
)≤ 0.
Theorem 1.22. Let Pi =[p(i)0 : · · · : p(i)n
], i = 1, . . . , g be points in Pn with
de�ning prime ideals ℘i, respectively, and let I(d) = ℘d1 ∩ · · · ∩ ℘dg be the ideal
of the scheme of fat points Xd = dP1 + . . .+dPg. Let Li = p(i)0 x0 + · · ·+p
(i)n xn,
for all i = 1, ..., g, and consider the power ideals Id = (Ld1, . . . , Lds).
Assuming g ≥ 2n, Xd gives independent conditions on hypersurfaces of any
degree for all d ≥ 1 if and only if Id is Hilbert generic for any d ≥ 1.
Proof. [⇒] Let Rd = S/Id for any d ≥ 1. Assume that I(e) impose independentconditions on hypersurfaces in any degree and for all e ≥ 1. Since Id is generatedin degree d, we look at the Hilbert function in degree i ≥ d. By ApolarityLemma, we have that, for all i ≥ d,
HF(Rd; i) = HF(I(i−d+1); i) = max
{0,
(i+ n
n
)− g(i− d+ n
n
)}.
Hence, we get, by Lemma 1.29,
HSRd(t) =
⌈1− gzd
(1− z)n+1
⌉=
⌈(1− zd)g
(1− z)n+1
⌉.
1Department of Mathematics, Stockholm University, Stockholm, SWEDEN.
E-mail address: ral�@math.su.se
29
[⇐] Assume that Ie is Hilbert generic, i.e. Re has Hilbert series equal to theseries given by Fröberg conjecture, Conjecture 1. By Lemma 1.29, we getHSRe
(t) =⌈(1− ze)m/(1− z)n+1
⌉=⌈(1− gze)/(1− z)n+1
⌉, for all e ≥ 1;
i.e.
HF(Re; i) =
{(i+nn
)if i < e,
max{
0,(i+nn
)− g(i−e+nn
)}if i ≥ e.
By Apolarity Lemma, for all i ≥ d,
HF(I(d); i) = HF(Ri−d+1; i) = max
{0,
(i+ n
n
)− g(d− 1 + n
n
)}.
Let S = C[x, y, z, t] be the polynomial rings in four variables and complexcoe�cients. Now, we want to prove that the Fröberg conjecture holds for 8generators of any degree d ≥ 1 in four variables. By Theorem 1.22, the questionis equivalent to showing that 8 fat points of multiplicity d in P3 in genericposition give independent conditions on hypersurfaces of any degree.
Theorem 1.23. Let P1, . . . , P8 be points in P3 in general position with de�ning
ideals ℘1, . . . , ℘8, respectively. For any d ≥ 1, let I(d) = ℘d1 ∩ . . . ∩ ℘d8 be the
de�ning ideal of the scheme of fat points Xd = dP1 + . . .+ dP8.
For any d ≥ 1, Xd imposes independent conditions on hypersurfaces of any
degree and the Hilbert series of the quotient S/I(d) is exactly
HSS/I(d)(t) =∑
0≤i≤2d−1
(i+ 3
3
)ti +
∑i≥2d
8
(d+ 2
3
)ti.
Proof. The property of imposing independent conditions on hypersurfaces ofcertain degree is an open condition on the scheme Xd. Indeed, it is equivalentto the maximality of the rank of the linear system obtained by evaluating allthe partial di�erentials of order ≤ d − 1 of a general hypersurface of certaindegree at the points P1, . . . , Pg. Consequently, it is enough to �nd an explicitarrangement of 8 points in P3 with the Hilbert series stated in the theorem.We consider the following set of points in P3:
P1 = [1 : 0 : 0 : 0], P2 = [0 : 1 : 0 : 0], P3 = [0 : 0 : 1 : 0], P4 = [0 : 0 : 0 : 1],
P5 = [1 : 1 : 1 : 0], P6 = [1 : 1 : 0 : 1], P7 = [1 : 0 : 1 : 1], P8 = [0 : 1 : 1 : 1].
Let Xd = dP1 + . . .+ dP8 with the de�ning ideal I(d).
Claim 1: I(d) is minimally generated in degree 2d with d+ 1 generators.
First, we show that Claim 1 implies the theorem. If I(d) is minimally generatedin degree 2d, for all i ≤ 2d− 1, the Hilbert function of the quotient equals thedimension of the whole space Si.
30
Moreover, we know that the Hilbert function of such a quotient is weaklyincreasing and eventually equals 8
(d+23
). Hence, assuming that in degree 2d the
ideal has dimension d+ 1, we have that
HF(S/I(d); 2d) =
(2d+ 3
3
)− (d+ 1) = 8
(d+ 2
3
),
and consequently it is equal to 8(d+23
)for all i ≥ 2d. This prove the theorem.
Let's prove Claim 1. We denote by ℘i the prime ideal of S corresponding tothe point Pi, e.g. ℘1 = (y, z, t). The ideal of our 0-dimensional scheme is
I(d) =
8⋂i=1
℘di .
We consider the following hyperplanes for any possible pairs of our variables:
Hxy = {x− y = 0}, . . . ,Hzt = {z − t = 0}.
Among such hyperplanes, we choose an independent 4-tuple, say
Hxy, Hxz, Hyt, Hzt.
We can easily check the following two properties:
� every hyperplane passes through exactly 4 of our points, e.g.
{P1, P2, P7, P8} ∈ Hzt;
� there are two pairs of such 4-uple with empty intersection along our Pi's.
Considering these two pairs, i.e. {Hxy, Hzt} and {Hxz, Hyt}, we can constructthe following two quadrics
C1 = HxyHzt, C2 = HxzHyt.
From the mentioned above properties of our hyperplanes, we may observe thateach of the two quadrics passes through all the P ′is exactly once and then,each power Cmi has singularities of degree m at each point; in other words, Cmibelongs to I(m) for any positive integer m. Hence, the following polynomialsare inside the degree 2d part of the ideal I(d),
Gd,0 = Cd1 ;
Gd−1,1 = Cd−11 · C2;
Gd−2,2 = Cd−21 · C22 ;
...
G0,d = Cd2 .
31
We want to show that these are the generators of the ideal in degree 2d.Firstly, we may observe that such polynomials are linearly independent. We
proceed by induction on the degree d. Assuming a linear combination αC1 +βC2 = 0, by intersecting with the hyperplane Hxy = {x− y = 0}, we get β = 0and, consequently, α = 0. Similarly, given a linear combination αC21 + βC1C2 +γC22 = 0, we get α = 0, by intersecting with the hyperplane Hxy, and γ = 0,by intersecting with Hxz; consequently, β = 0. With these two base steps, wecan do the induction. Let's take a linear combination
∑i=0,...,d αiGd−i,i = 0.
By intersecting with the planes Hxy and Hxz, we get α0 = 0 and αd = 0,respectively. Consequently, we get a linear combination,
C1C2 ·d−2∑j=0
αj+1Cd−2−j1 Cj2 = 0;
by induction, we are done.
After showing that the above polynomials are linearly independent, we nowhave to prove that they are enough to generate the whole degree 2d part of theideal, namely that the dimension of the linear system L2d(Xd) is equal to d+ 1and not bigger. We proceed by induction on the degree d.
The case d = 1 is trivial. For the general case, we look at the linear systemL2d(Xd + P ) of hypersurfaces of degree 2d through our scheme Xd plus oneextra simple point P .
Claim 2: the dimension of L2d(Xd + P ) is equal to d.
Assuming Claim 2, we would have that the linear system L2d(X) has theright dimension; indeed
dClaim 2
= dimC L2d(Xd + P ) ≥ dimC L2d(Xd)− 1 ≥ (d+ 1)− 1.
Let's prove Claim 2. The aim is to show that, choosing point P in a cleverway, the quadric given by the two plane Hxy and Hzt is a �xed component forall the surfaces of degree 2d through the scheme X(d) +Q; thus, by induction,
dimC L2d(Xd +Q) = dimC L2d−2(Xd−1) = d.
The symmetry of the problem, suggests we choose the point Q on the inter-section line l = {x−y = z−t = 0} between the two planes. Let's take, for exam-ple, a point on the line with coordinatesQm := [2m+1 : 2m+1 : 2m+3 : 2m+3]for some m ≥ 0.
On the plane Hxy we consider the coordinates {x, z, t} and we consider thegeneral conic with equation
α1x2 + α2xz + α3xt+ α4z
2 + α5zt+ α6t2 = 0;
imposing the passage through the points P3, P4, P5, P6 on the plane, we get thefollowing conditions
α4 = α6 = 0, α1 + α2 = α1 + α3 = 0,
32
thus, we get the pencil of conics
α1x2 − α1xz − α1zt− α5zt = 0.
Imposing the passage through the point Qm, we get the condition
α5 =2(2m+ 1)(2m+ 3)− (2m+ 1)2
(2m+ 3)2α1 =
(2m+ 1)(2m+ 5)
(2m+ 3)2α1,
and then the unique conic
Q = (2m+ 3)2x2 − (2m+ 3)2xz − (2m+ 3)2xt+ (2m+ 1)(2m+ 5)t2 = 0.
Intersecting Q with the intersection line l, i.e. imposing z − t = 0, we get theequation
(2m+ 3)2x2 − 2(2m+ 3)2xz + (2m+ 1)(2m+ 5)t2 = 0
=⇒ [(2m+ 3)x− (2m+ 1)t] [(2m+ 3)x− (2m+ 5)t] = 0.
Thus, the conic Q intersects the line l also in two distinct points: the point Qmand also Rm = [2m+ 5 : 2m+ 5 : 2m+ 3 : 2m+ 3].
We play the same game on the plane Hzt considering the point Rm. Impos-ing the passage through the points P1, P2, P7, P8 to the conics in the variables{x, y, z}, we get the pencil of conics
β2xy − β6xz − β6yz + β6z2 = 0.
Imposing the passage through the point Rm, we get the condition
β2 =2(2m+ 5)(2m+ 3)− (2m+ 3)2
(2m+ 5)2β6 =
(2m+ 7)(2m+ 3)
(2m+ 5)2β6;
thus, we get a unique conic intersecting the line l at the two distinct points Rmand Qm+2. Hence, we come back on the plane Hxy and we continue the samegame starting from the point Qm+2, and so on.
H
H
xy
P
P P
P
PP
PP
R
3
4 5
6
71
28
zt
m
m+1m
33
In this way, we can �nd on the two planes in�nitely many conics whichhave to be �xed component for our linear system; thus, the two planes are�xed components and the Claim 2 is proved.
To complete the proof of the theorem, we just need to prove that thereare no elements in the degree 2d − 1 part of the ideal I(Xd); indeed, assumeF ∈ I(Xd) of degree 2d−1; thus, xF ∈ I(Xd). Consequently, there exists a nulllinear combination between xF and the Gi's. By specializing on the hyperplanex = 0 and using the linear independence of the Gi's, we are done.
Corollary 1.24. The Fröberg conjecture holds in four variables and for generic
ideals with 8 generators of the same degree d.
Proof. Directly from Theorem 1.22 and Theorem 1.23
Betti numbers
The study of the Hilbert function of 8 generic fat points in P3 gave us the proofof the fact that the Fröberg conjecture holds for such a numerical character.However, since in the proof of Theorem 1.23 we have computed explicitly a set ofgenerators for the corresponding ideal, we can go further in our computations.
First, let's recall the basic de�nitions about Betti numbers.
Given a �nitely generated S-module M , there exists a �nitely generatedfree S-module F0 and a surjective homomorphism ϕ0 mapping the generatorsof F0 to a set of generators of M . The kernel of that homomorphism is again a�nitely generated S-module and we have an exact sequence of S-modules
0 // kerϕ0// F0
ϕ0 // M // 0
Now, we can �nd another surjective homomorphism ϕ1 mapping the generatorsof a free �nitely generated S-module F1 to the generators of kerϕ0, i.e. therelations among the generators of F0. Such relations are called the syzygies ofthe module. Then we continue this procedure considering the kernel of ϕ1 andso on. Hence, we get a free resolution of our S-module M , i.e a long exactsequence
F• : . . . // F2// F1
// F0// M // 0.
By the Hilbert's syzygy theorem, we also know that, for any M �nitelygenerated module over a polynomial ring S, there is a �nite free resolution, i.e.there are F1, . . . , Fn �nitely generated free S-modules such that
F• : 0 // Fn // . . . // F1// F0
// M // 0.
Of course, if we want to optimize this construction, we can start by consideringa minimal set of generators of our moduleM and then, at each step, a minimalset of generators for the module of syzygies. In that way, we get a minimal free
resolution F• for our module M . The length of the minimal free resolution n iscalled the projective dimension of the moduleM and denoted by pdim(M).
34
Moreover, if we have also a graded structure on M , we can even assume tohave graded maps of degree 0 in such a minimal resolution. In that way, weget a set of invariants for the module M . Indeed, considering a minimal freeresolution with degree 0 maps, we can write, for all i = 1, . . . , n,
Fi =⊕j∈Z
S(−j)βi,j .
The integers βi,j are called the graded Betti numbers of M . Such numberscan be seen as homological quantities since one can prove that
βi,j = dim[TorSi (M,C)]j ;
this fact shows that such numbers depends only on the module M and not onthe minimal resolution.1
Often, the Betti numbers are represented as Betti table, like the following
β(M) =
. . . i . . ....
...j . . . βi,i+j . . ....
...
Remark 1.25. In the entry (i, j) of the table we put the Betti number βi,i+jby minimality of the resolution which de�nes them. Moreover, the minimalityof the resolution gives us another piece of useful information about the Bettinumbers and the shape of the Betti table, namely
min{j | βi+1,j 6= 0} > min{j | βi,j 6= 0}.
Moreover, because of the relation between exact sequences of S-modules withdegree 0 maps and the Hilbert function of the modules in the sequence, theBetti numbers allow us to write the Hilbert series of a module M in a moreexplicit way. Hence, given an S-module M and its Betti numbers (βi,j), weobtain
HSM (t) =
∑pdim(M)i=0
∑j∈Z(−1)iβi,jt
j
(1− t)n+1.
Now, we want to reconsider the case of 8 fat points in generic position in P3
and compute their Betti numbers.
Let's take R(d) = S/I(d) where I(d) is the ideal of the 0-dimensional schemeX = dP1 + . . . + dP8 where Pi's are points in general position in P3 and d apositive integer. We have proved in Theorem 1.23 that
HSR(d) =
2d−1∑j=0
(3 + j
j
)zj + 8
(d+ 2
3
)z2d
1− z.
1See Eisenbud [18] for basic notions in Homological Algebra.
35
Since R(d) is a Cohen-Macaulay ring, we can take a NZD L of degree 1. Con-sequently, from the exact sequence given by the multiplication by L, we havethat
HSR(d)/(L)(z) = (1− z) HSR(d)(z) =
= (1− z)2d−1∑j=0
(3 + j
j
)zj + 8
(d+ 2
3
)z2d;
in particular, [R(d)/(L)]2d+1 = 0 and, since L is a NZD of R(d), we have that
βi,j = dim TorSi,j
(R(d),C
)= dim TorSi,j
(R(d)/(L),C
).
In order to compute the last Tor, one can start from a resolution of C,which is given by the Koszul complex K•, and then tensoring by R(d)/(L),let's call such tensor product K•. Now, by de�nition of the Koszul complex,which is given by the exterior algebra
∧C4, we have that Ki = 0 for all i ≥ 5.
Otherwise, we have that the elements of the graded part Ki,j of Ki consist oflinear combinations of elements of type f · tj1 · · · tji , where deg(f) = j − i andwhere {t1, . . . , t4} is a basis of C4. Thus, since [R(d)/(L)]2d+1 looking at thehomology, we get TorSi,j(R
(d)/(L)) = 0 for all j ≥ i+ 2d+ 1. Thus, we can onlyhave a few possibly non-zero Betti numbers, namely
0 1 2 3 4 50 1 - - - - -
2d-1 - β1,2d β2,2d+1 β3,2d+2 β4,2d+3 -2d - β1,2d+1 β2,2d+2 β3,2d+3 β4,2d+4 -
2d+1 - - - - - -
Lemma 1.26. There are no linear syzygies among the generators of our ideal
I(d), i.e. β2,2d+1 = 0.
Proof. Since the vanishing of a Betti number is equivalent to the vanishing ofthe corresponding Tor, it is given by the surjectivity or the injectivity of somelinear map, namely by the maximality of the rank of some matrix. Hence, it isan open condition and then is again enough to show that such condition holdsfor a speci�c example in order to get the result for the generic case.
Let's consider the same arrangement of eight fat points as in Theorem 1.23.We have proved that the corresponding ideal is generated by d+ 1 polynomialsGd−i,i := Cd−i1 Ci2, for i = 0. . . . , d, where C1 = (x − y)(z − t) and C2 = (x −z)(y − t).
We proceed by induction over d. First, assume that there is a linear syzygy,i.e. there exists L0, L1 linear forms such that
L0G1,0 + L1G0,1 = L0(x− y)(z − t) + L1(x− z)(y − t) = 0.
Thus, all the third order derivatives of such polynomial should be equal tozero. Since the monomials xz and yt appear only in G1,0 and not in G0,1,
36
the derivatives ∂x2z, ∂y2t, ∂xz2 , ∂yt2 give us L0 = 0. Similarly, by using themonomials xy and zt, we get L1 = 0.
Let's now assume that there is a linear syzygy in the case d = 2, i.e. there existlinear forms L0, L1 and L2 such that
L0G2,0 + L1G1,1 + L2G0,2 = 0.
Again, monomials x2z2 and y2t2 appear only in G2,0, thus, imposing that thedegree 5 derivatives ∂x3z2 , ∂y3t2 , ∂x2z3 , ∂y2t3 vanish, we get L0 = 0. Similarly, byconsidering monomials x2y2 and z2t2, which appear only in G0,2, and deriva-tives ∂x3y2 , ∂x2y3 , ∂z3t2 , ∂z2t3 , we get L2 = 0. Consequently, L1 = 0.
In general, assuming we have a linear syzygy
d∑i=0
LiGd−i,i =
d∑i=0
LiCd−i1 Ci = 0,
we can consider the monomials xdzd and ydtd which appear only in Gd,0. Byimposing the vanishing of all the derivatives ∂xd+1zd , ∂yd+1td , ∂xdzd+1 , ∂ydtd+1 ,we get L0 = 0. Similarly, we can use the monomials xdyd and zdtd to concludeLd = 0. Thus, we can rewrite the linear syzygy as
C1C2 ·d−2∑j=0
lj+1Cd−2−j1 Cj2 = 0,
and then use induction.
Now, since there are no linear syzygies in our ideal I(d), we have thatβ2,2d+1 = 0 and consequently, by basic property of Betti numbers, also β3,2d+2 =β4,2d+3 = 0. Thus, we can read the remaining Betti numbers directly from theHilbert series of R(d) since it is of the form
1− β1,2dz2d − β1,2d+1z2d+1 + β2,2d+2z
2d+2 − β3,2d+3z2d+3 + β4.2d+4z
2d+4
(1− z)4.
By our computation of HSR(d)(z) in Theorem 1.23, we get
(1− z)4 HSR(d) = (1− z)42d−1∑j=0
(3 + j
j
)zj + (1− z)38
(d+ 2
3
)z2d. (1.5)
From that, we already can see that β4,2d+4 = 0. Now, one can show that
(1− z)42d−1∑j=0
(3 + j
j
)zj = 1 +
3∑j=0
(−1)j+1
(3
j
)(3 + 2d
2d
)2d
2d+ jz2d+j .
The idea to prove such an equality is to show that both sides have the samederivatives, see Fröberg and Löfwal [23, Proposition 5.2] for a similar compu-tation. Thus, we can continue from equation (1.5), getting
(1− z)4 HSR(d) = 1 +
3∑j=0
(−1)j+1
(3
j
)[(3 + 2d
2d
)− 8
(d+ 2
3
)]z2d+j .
37
By direct computations, we �nally get
(1− z)4 HSR(d) = 1− (d+ 1)z2d−4
(d+ 1
2
)z2d+1 +d(4d+ 5)z2d+2−4
(d+ 1
2
).
Hence, the Betti diagram of R(d) is as follows
0 1 2 3 40 1 - - - -
2d-1 - d+ 1 - - -
2d - 4(d+12
)d(4d+ 5) 4
(d+12
)-
2d+1 - - - - -
Remark 1.27. Having settled Theorem 1.2, it is reasonable to try to gener-alize the result by considering, in the projective space Pn, the 0-dimensionalschemes supported on the 2n points with an odd number of 1's and 0's as theircoordinates. Unfortunately, already the n = 4 case is much more di�cult. A�rst di�erence is that, from computer experiments with Macaulay2, we �guredout that the ideal of fat points can be generated in more than one degree, asopposed to the n = 3 case. Moreover, the generators do not look as nice as inthe case we have considered, where they are simply products of linear forms.
Section 1.3
A class of power ideals (joint work with J. Backelin 1)
In this section, we want to consider a special class of power ideals dependingon three positive indices and recently introduced in connection with a War-ing problem for polynomial rings in Fröberg, Ottaviani, and Shapiro [25]. Wediscuss the Waring problem in more details in the next chapter.
For any triple (n, k, d) of positive integers, �xed ξ a primitive kth-root ofunity, we consider the homogeneous ideal In,k,d generated by the kn powers(x0 + ξg1x1 + . . .+ ξgnxn)(k−1)d where 0 ≤ gj ≤ k − 1 for all j = 1, . . . , n. Wedenote the quotient ring as Rn,k,d := C[x0, . . . , xn]/In,k,d and with [Rn,k,d]j itshomogeneous component of degree j.
1.3.1 Multicycle gradation
Let Zk = {[0]k, [1]k, . . . , [k− 1]k} be the cyclic group of integers modulo k. Letξ be a primitive kth-root of unity and observe that, for any ν ∈ Zk, the com-plex number ξν is well-de�ned. We will usually use a small abuse of notationdenoting a class of integer modulo k simply with its representative between 0and k − 1; e.g. when we will consider the scalar product between two vectors
1Department of Mathematics, Stockholm University. Stockholm, SWEDEN.
E-mail address: [email protected]
38
g,h ∈ Zn+1k , denoted by 〈g,h〉, we will mean the usual scalar product consid-
ering each entry of the two vectors as the smallest positive representative ofthe corresponding class.
Consider, for each g = (g0, . . . , gn) ∈ Zn+1k , the polynomial
φg :=
(n∑i=0
ξgixi
)D, where D := (k − 1)d.
Hence, In,k,d is by de�nition the ideal generated by all φg, with g ∈ 0×Znk . It ishomogeneous with respect to the standard gradation, but it is also homogeneouswith respect to the Zn+1
k -gradation we are going to de�ne.
Consider the projection πk : N −→ Zk given by πk(n) = [n]k. For any vectora = (a0, . . . , an) ∈ Nn+1, we de�ne the multicyclic degree as follows.
Given a monomial xa := xa00 . . . xann , we set
mcdeg(xa) := πn+1k (a) = ([a0]k, . . . , [an]k).
Thus, combining this multicyclic degree with the standard gradation, we getthe multigradation on the polynomial ring S given by
S =⊕i∈N
Si =⊕i∈N
⊕g∈Zn+1
k
Si,g, where Si,g := Si ∩ Sg;
where, for any i1, i2 ∈ N and g1,g2 ∈ Zn+1k , we have that
Si1,g1 · Si2,g2 = Si1+i2,g1+g2 .
Remark 1.28. For 0 := (0, . . . , 0), we get obviously that S0 = C[xk0 , . . . , xkn],
and then, for any i ∈ N,Si,0 6= 0 if and only if i = jk for some j ∈ N,
in such case
dimC Sjk,0 =
(n+ j
n
).
For any arbitrary multicycle g = (g0, . . . , gn) ∈ Zn+1k , we de�ne the partition
vector to be part(g) := (#{gi = 0}, . . . ,#{gi = k − 1}) and the weight of gas wt(g) :=
∑nj=0 gj . Clearly, the weight is non-negative and
wt(g) = 0 if and only if g = 0.
Lemma 1.29. Let i ∈ N and g ∈ Zn+1k . Then,
Si,g 6= 0 if and only if i− wt(g) = jk, for some j ∈ N.In such case,
dimCSi,g =
(n+ j
n
).
39
Proof. Given a monomial xa with i = deg(xa), consider g = πn+1k (a). Hence,
we have that xa−g ∈ Si−wt(g),0. Hence,
dimC Si,g = dimC Si−wt(g),0 =
(n+ j
n
).
Now, we denote with Gk,n,i the set set of all multicycles satisfying the twoequivalent conditions of Lemma 1.29, i.e.
Gk,n,i := {h ∈ Zn+1k | i− wt(h) ∈ kN} = {h ∈ Zn+1
k | Si,h 6= 0}.
Coming back to our ideal, since we can write SD =⊕
g∈Zn+1k
SD,g, we can
represent the generator φ0 = (x0 + . . .+ xn)D of In,k,d as
φ0 =∑
g∈Zn+1k
ψg, where ψg ∈ SD,g.
Clearly, if ψg 6= 0 then g ∈ Gk,n,D, but one can also check that actually
ψg 6= 0 ⇐⇒ g ∈ Gk,n,D.
In particular, under the equivalent latter conditions, we have that,
ψg =∑
d0+...+dn=Dπn+1k (d0,...,dn)=g
(D
d0, . . . , dn
)xd.
With the following example, we make this construction more explicit.
Example 1.30. Consider the case k = 2, n = 2, d = 4 and φ0 = (x0+x1+x2)4.We have
ψ(0,0,0) = x40 + 6x20x21 + 6x20x
22 + x41 + 6x21x
22 + x42;
ψ(1,0,0) = ψ(0,1,0) = ψ(0,0,1) = ψ(1,1,1) = 0;
ψ(1,1,0) = 4x30x1 + 12x0x1x22 + 4x0x
31;
ψ(1,0,1) = 4x30x2 + 12x0x21x2 + 4x0x
32;
ψ(0,1,1) = 4x1x32 + 12x20x1x2 + 4x1x
32.
We can notice that, since (1, 0, 0) 6∈ G2,2,4, we already expected ψ(1,0,0) = 0,and similarly for (0, 1, 0), (0, 0, 1) and (1, 1, 1).
Lemma 1.31. For any g ∈ Zn+1k , one has
φg =∑
h∈Gk,n,D
ξ〈g,h〉ψh;
conversely,
ψg = k−n−1∑
h∈Zn+1k
ξ−〈g,h〉φh.
40
Proof. From the de�nition, we can write
φg =
(n∑i=0
ξgixi
)D=
∑d0+...+dn=D
(D
d0, . . . , dn
) n∏l=0
ξgldlxdll =
=∑
d0+...+dn=D
(D
d0, . . . , dn
)ξ〈g,d〉xd.
Now, we can consider for each d = (d0, . . . , dn) the vector πn+1k (d) = h ∈ Zn+1
k .Since ξ is a kth root of unity, we have ξ〈g,d〉 = ξ〈g,h〉. Thus,
φg =∑
h∈Gk,n,D
ξ〈g,h〉∑
d0+...+dn=Dπn+1k (d)=h
(D
d0, . . . , dn
)xd =
∑h∈Gk,n,D
ξ〈g,h〉ψh.
For the second part of the statement, we consider the following equality whichfollows from the �rst part already proved. For any m ∈ Zn+1
k ,∑g∈Zn+1
k
ξ−〈g,m〉φg =∑
g∈Zn+1k
∑h∈Gk,n,D
ξ〈g,h〉−〈g,m〉ψh.
On the right hand side, we have{if m = h :
∑g∈Zn+1
kψh = kn+1ψh;
if m 6= h :∑
g∈Zn+1k
ξ〈g,h−m〉ψh =∑
g∈Zn+1k
ξg00 . . . ξgnn ψh = 0.
Hence, we have the set {ψg}g∈Gk,n,Dof nonzero polynomials with distinct mul-
ticyclic degree and consequently linearly independent. In other words, we haveproved the following proposition.
Proposition 1.32. In,k,d is minimally generated by {ψg}g∈Gk,n,D.
Theorem 1.33. The cardinality of Gk,n,D is given by
|Gk,n,D| =
=∑i≥0
∑ν2,...,νk−1≥0
(n+ 1
D − ki−∑k−1j=1 (j − 1)νj
)(D − ki−
∑k−1j=1 (j − 1)νj
ν2, . . . , νk−1, D −∑k−1j=2 jvj
)=
=∑
i,ν2,...,νk−1≥0
(n+ 1
ν2, . . . , νk−1, D − ki−∑k−1j=2 jνj , n+ 1−D + ki+
∑k−1j=2 (j − 1)νj
).
In particular, if k = 2, then this number of generators equals∑i≥0(n+1d−2i
).
41
Proof. For any g ∈ Gk,n,D, we can write ψg = f(xk0 , . . . , xkn)xg where f is a
homogeneous polynomial of degree i and part(g) = (0, ν1, . . . , νk−1).To count the number of elements of Gk,n,D, we have
( n+1D−ki−
∑k−1j=1 (j−1)νj
)ways
to choose the variables with nonzero exponent modulo k and, for each such
choice, there are( D−ki−
∑k−1j=1 (j−1)νj
ν2,...,νk−1,D−∑k−1
j=2 jvj
)ways to distribute the exponents.
Example 1.34. For k = 4, d = 3, n = 2 we get that the number of minimalgenerators is
(3
0,3,0,0
)+(
30,1,2,0
)+(
31,1,0,1
)+(
32,0,1,0
)+(
30,0,1,2
)= 16. This means
that the original generators φg are linearly independent.
Theorem 1.35. If k = 2, the generators {φg}g∈0×Zn2are linearly independent
if and only if n+ 1 ≤ d.
Proof. {ψg} is linearly independent, and they are∑i≥0(n+1d−2i
)many. This sum
equals 2n if and only if n+ 1 ≤ d.
1.3.2 Hilbert function of the power ideal In,k,d
In order to simplify the notation, when there will be no ambiguity, we willdenote I := In,k,d and R := Rn,k,d = S/I with the multicycling gradationdescribed in the previous section R =
⊕i∈N⊕
g∈Zn+1k
Ri,g.
De�nition 1.36. For 0 ≤ i ≤ d and given a vector h ∈ Zn+1k , we de�ne the
map
µi,h : Di,h :=⊕
g∈Zn+1k
Si,h−g −→ Si+D,h,
(. . . , fg, . . .) 7−→∑
g∈Zn+1k
fgψg.
given by the multiplication by each ψg ∈ SD,g.
Remark 1.37. In order to work with relevant examples, we'll assume alwaysthat i + D − wt(h) ∈ kZ in order to have Si+D,h 6= 0. We may also observethat, under such assumption, we have the following equivalence
i− wt(h− g) ∈ kZ⇐⇒ D − wt(g) ∈ kZ;
in other words, again from the properties of this multicyclic gradation explainedin the previous section, we have
Si,h−g 6= 0⇐⇒ ψg 6= 0.
Thus, it makes sense to study the injectivity of the µi,h's and it will be thecrucial step for our computations.
Lemma 1.38. Given 0 ≤ i ≤ d and h ∈ Zn+1k , if i + D − wt(h) ∈ kN and
wt(h) ≤ (k − 1)(d− i), we have
dim(Di,h) ≤ dim(Si+D,h);
with equality if wt(h) = (k − 1)(d− i).
42
Proof. In such numerical assumptions, we have that Di,h is simply Si; thus,
dimCDi,h =
(n+ i
n
);
moreover, we may observe that, for some integer m ≥ 0,
km = i+D − wt(h) ≥ i+D − (k − 1)(d− i) = ki;
hence, i+D − wt(h) = k(i+ j) for some j ≥ 0 and
dim(Si+D,h) =
(n+ i+ j
n
).
For any 0 ≤ i ≤ d and h ∈ Zn+1k , the image of the map µi,h is simply the part
of multicycling degree (i,h) of our ideal I. These maps will be the main toolin our computations regarding the Hilbert function of I and its quotient ringR. By Remark 1.45 and Lemma 1.38, it makes sense to ask if µi,h is injectivewhenever wt(h) ≤ (k − 1)(d − i) and i + D − wt(h) ∈ kZ: in that cases, thedimension of Ii+D,h in degree i will be simply the dimension of Di,h = Si. Onthe other hand, again by Lemma 1.38, one could hope that µi,h is surjective inall the other cases to get, consequently, Ri+D,h = 0.
This is true for k = 2 as we are going to prove in the next section.
The k = 2 case
In this case, D = (k − 1)d = d. Moreover, as we said in Remark 1.45, we'llconsider only the maps µi,h such that i+ d− wt(h) is even.Lemma 1.39. In the same notation as above, we have:
1. µd,0 is bijective;
2. µi,h is injective if wt(h) ≤ d− i;
3. µi,h is surjective if wt(h) ≥ d− i.
Proof. (1) The map µd,0 is surjective because Theorem 2.11. It is the mainresult in Fröberg, Ottaviani, and Shapiro [25] and, because of the relationwith Waring's problem, we discuss it in the next chapter. The map µd,0 isalso injective because we are in the limit case of Lemma 1.38, i.e. where thedimensions of the source and the target are equal.
(2) Given a monomial M with M ∈ Sd+i,h, there exists a monomial M ′
such that MM ′ ∈ S2d,0; indeed, it is enough to consider the monomial xh toget mcdeg(xhM) = 0 and then we can multiply for any monomial with theright degree to get degree equal to 2d and multicyclic degree equal to 0. Hence,the injectivity of µi,h follows from (1).
(3) If wt(h) = (d−i), we are in the limit case of Lemma 1.38 and then, frominjectivity of µi,h, it follows also the surjectivity. Instead, the case wt(h) >
43
(d − i) follows from the previous one because, given any monomial M withM ∈ Sn,h and n − wt(h) = 2m, then M is a product of a monomial M ′ withM ′ ∈ Sn−2m,h.
We are now able to study the Hilbert function of our class of power ideals andtheir quotients.
Lemma 1.40. In the same notation as above, we have:
1. if i < d, Ii = 0;
2. if i = j + d with j ≥ 0, Ri,h 6= 0 if and only if
h ∈ Hj := {h′ | i− wt(h′) ∈ 2N, wt(h′) < d− j, wt(h′) ≤ n+ 1};
moreover, if h ∈ Hj, then
dimCRi,h = dimC Si,h −(n+ j
n
).
Proof. Since I has generators in degree d, then Ii = 0 for all i < d. Considernow i = d + j for some j ≥ 0. Since Ri =
⊕h∈Zn+1
kRi,h, we will focus on
the dimension of each summand Ri,h. Fix h ∈ Zn+1k . We have seen that I =
(ψg | g ∈ G2,n,D); hence, Ii,h = Im(µj,h). By Lemma 1.39, for wt(h) ≥ d − j,we know that µj,h is surjective and then Ii,h = Si,h; consequently, Ri,h =0. Moreover, by Lemma 1.29, we need to consider only h ∈ Zn+1
k such thati − wt(h) ∈ 2N otherwise Si,h = 0 and consequently, Ri,h = 0. Thus, we justneed to consider h in the set Hj de�ned in the statement. By Lemma 1.39, inthat numerical assumptions, µj,h is injective and then
dimC Ii,h =∑
g∈Zn+1k
dimC Sj,h−g = dimC Sj =
(n+ j
n
),
or equivalently,
dimCRi,h = dimC Si,h −(n+ j
n
).
Theorem 1.41. The Hilbert function of the quotient ring R is given by:
1. if i < d, HF(R; i) =(n+in
);
2. if i = j + d with j ≥ 0,
HF(R; i) =∑h∈Hj
dimCRi,h =∑
h<d−ji−h∈2N
(n+ 1
h
)((n+ i−h
2
n
)−(n+ j
n
))
44
Proof. For i < d it is trivial. Hence, consider i = j + d with j ≥ 0. First, wemay observe that, by Lemma 1.40, whenever h ∈ Hj , the dimension of Ri,hdepends only on the weight of h. Indeed, considering h ∈ Hj and denotingh := wt(h), we get, by Lemma 1.29,
dimCRi,h = dimC Si−h,0 −(n+ j
n
)=
(n+ i−h
2
n
)−(n+ j
n
).
To conclude our proof, we just need to observe that, �xed a weight h, we haveexactly
(n+1h
)vectors h ∈ Zn+1
2 with such weight.
Corollary 1.42. R2d−1 = 0.
Proof. R2d−1,h 6= 0 if and only if wt(h) is odd and wt(h) < 1, so never.
In the following example, we explicit our algorithm in a particular case in orderto help the reader in the comprehension of the theorem.
Example 1.43. Let's take n + 1 = 4, i.e. S = C[x0, . . . , x3] and d = 5. Wecompute the Hilbert function of the quotient R = S/I2,3,5 where
I2,3,5 =((x0 ± x1 ± x2 ± x3)5
).
For i < 5, we have
HF(R; i) =
(3 + i
3
).
For i = 5 (j = 0), we have that H0 = {h | wt(h) = 1, 3}, hence
HF(R; 5) =∑
wt(h)=1
dimCR5,h +∑
wt(h)=3
dimCR5,h =
=
(4
1
)(dimC(S4,0)− 1) +
(4
3
)(dimC(S2,0)− 1) =
= 4(10− 1) + 4(4− 1) = 36 + 12 = 48.
For i = 6 (j = 1), we have that H1 = {h | wt(h) = 0, 2}, hence
HF(R; 6) = dimCR6,0 +∑
wt(h)=2
dimCR6,h =
= (dimC(S6,0)− 4) +
(4
2
)(dimC(S4,0)− 4) =
= (20− 4) + 6(10− 4) = 16 + 36 = 52.
For i = 7 (j = 2), we have that H2 = {h | wt(h) = 1}, hence
HF(R; 7) =∑
wt(h)=1
dimCR7,h =
(4
1
)(dimC(S6,0)− 10) = 4(20− 10) = 40.
45
For i = 8 (j = 3), we have that H3 = {0}, hence
HF(R; 8) = dimCR8,0 = dimC(S8,0)− 20 = 35− 20 = 15.
For i ≥ 9 (j ≥ 4), we can easily see that Hj = ∅. Thus, the Hilbert functionis
i 0 1 2 3 4 5 6 7 8 9HF(R; i) 1 4 10 20 35 48 52 40 15 -
With the following theorem, we are going to work on our result in order tocompute more explicitly the Hilbert series in cases with small number of vari-ables.
Theorem 1.44. The Hilbert series of R2,1,d is given by (1−2td+ t2d)/(1− t)2.The Hilbert series of R2,2,d, for d ≥ 2 is given by
HS(R2,2,d; t) =
(1− 4td + dt2d−1 + 3t2d − dt2d+1
)(1− t)3
=
=
d−1∑i=0
(i+ 2
2
)ti +
d−2∑i=0
((d+ i+ 2
2
)− 4
(i+ 2
2
))td+i.
The Hilbert series of R2,3,d, for d ≥ 3 is given by
HS(R2,3,d; t) =
=
(1− 8td +
(d2
)t2d−2 + 4dt2d−1 − (d2 − 7)t2d − 4dt2d+1 +
(d+12
)t2d+2
)(1− t)4
=
=
d−1∑i=0
(i+ 3
3
)ti +
d−3∑i=0
((d+ i+ 3
3
)− 8
(i+ 3
3
))td+i +
(d+ 1
2
)t2d−2.
Proof. Case n+ 1 = 2. Simply, we have a complete intersection and it followsthat the Hilbert series is (1− 2td + t2d)/(1− t)2.
Case n+ 1 = 3. From Lemma 1.39, we have that [I2,2,d]d+j = Sj [I2,2,d]d forany 0 ≤ j ≤ d− 3 since wt(h) ≤ d− 3 for all possible h. Since 2d− 2 is even,we get that wt(h) should be even and then, wt(h) ≤ 2 = d − (d − 2); thus,we get injectivity also in this degree. Now, from Theorem 1.41, we get thatdimC([R2,2,d]d+j) = dimC(Sd+j)−#(Hj) ·
(n+jn
). In our numerical assumption,
it is clear that, for 0 ≤ i ≤ d − 3, Hi is exactly the half of all possible vectorsin Zn+1
2 , i.e. #(Hi) = 2n; hence,
HS(R2,2,d; t) =
d−1∑i=0
(i+ 2
2
)ti +
d−2∑i=0
((d+ i+ 2
2
)− 4
(i+ 2
2
))td+i.
A simple calculation shows that (1 − t)3 HS(R2,2,d; t) = (1 − 4td + dt2d−1 +3t2d − dt2d+1).
46
Case n+1 = 4. From Lemma 1.39, since wt(h) ≤ 4 for all possible h, we getthat [I2,3,d]d+i = Si[I2,3,d]d for all 0 ≤ i ≤ d− 4. Moreover, since 2d− 3 is odd,we get that wt(h) should be odd and consequently wt(h) ≤ 3 = d − (d − 3);hence, we have injectivity also in this degree. Moreover, for all 0 ≤ i ≤ d − 3,we get that Hi is half of all possible vectors in Zn+1
2 , i.e. Hi has cardinalityequal to 2n. Now, we just miss to compute the dimension of [R2,3,d]2d−2. Byde�nition, the vectors h ∈ Hd−2 have to be odd, since 2d − 2 is odd, and tosatisfy the condition wt(h) < 2; thus, we get only h = 0 and #(Hd−2) = 1.Thus, by Theorem 1.41,
dimC([R2,3,d]2d−2) = dimC([R2,3,d]2d−2,0) = dimC(S2d−2,0)−(
3 + d− 2
3
)=
=
(d+ 2
3
)−(d+ 1
3
)=
(d+ 1
2
).
Putting together our last observations, we get
HS(R2,3,d; t) =
=
d−1∑i=0
(i+ 3
3
)ti +
d−3∑i=0
((d+ i+ 3
3
)− 8
(i+ 3
3
))td+i +
(d+ 1
2
)t2d−2.
A simple calculation shows that
(1− t)4 HS(R2,3,d; t) =
= 1− 8td +
(d
2
)t2d−2 + 4dt2d−1 − (d2 − 7)t2d − 4dt2d+1 +
(d+ 1
2
)t2d+2.
Remark 1.45. From the proof of Theorem 1.44, we can say something morealso about the Hilbert series of R2,n,d, even for more variables.
Assuming d ≥ n, by using the same ideas as in the theorem above, we getthat for all 0 ≤ j ≤ d−n, the (d+ j)th-coe�cient of our Hilbert series is equalto
HF(R2,n,d; d+ j) =
(n+ d+ j
n
)− 2n
(n+ j
n
).
Moreover, we get that, for any d ≥ 2, Hd−2 = {0} and consequently,
HF(R2,n,d; 2d− 2) = dimC([R2,n,d]2d−2) = dimC([R2,n,d]2d−2,0) =
= dimC(S2d−2,0)−(n+ d− 2
n
)=
(n+ d− 1
n
)−(n+ d− 2
n
)=
=
(n+ d− 2
n− 1
).
47
Similarly, we have that, for any d ≥ 3, Hd−3 = {h ∈ Zn+1k | wt(h) = 1}, thus
HF(R2,n,d; 2d− 3) = dimC([R2,n,d]2d−3) =∑
wt(h)=1
dimC([R2,n,d]2d−3,h) =
= (n+ 1)
[dimC(S2d−2,0)−
(n+ d− 2
n
)]=
= (n+ 1)
(n+ d− 2
n− 1
).
Conjecture 2. R2,n,d is level algebra, i.e. Soc(R2,n,d) = [R2,n,d]2d−2. If so,
from Remark 1.45, we would have that Soc(R2,d,n) has dimension(n+d−2n−1
).
The k > 2 case.
We would like to generalize our results for the cases k > 2. Inspired byLemma 1.38, we conjecture the following behavior of the maps µi,h.Conjecture 3. In the same notation as De�nition 1.36, we have
1. µi,h is injective if wt(h) ≤ (k − 1)(d− i);
2. µi,h is surjective if wt(h) ≥ (k − 1)(d− i).
Following the same ideas as Lemma 1.40, from Conjecture 3 we would get thefollowing results.
Conjecture 4. In the same notation as above, we have
if i = j +D with j ≥ 0, Ri,h 6= 0 if and only if
h ∈ Hj := {h′ | i− wt(h′) ∈ kN, wt(h′) < d− j, wt(h′) ≤ (k − 1)(n+ 1)};
moreover, if h ∈ Hj, then
dimCRi,h = dimC(Si,h)−(n+ j
n
).
Proposition 1.46. Conjecture 3 =⇒ Conjecture 4.
Proof. Follow the proof of Theorem 1.41.
Remark 1.47. From these conjectures, it would follow a direct generalizationof the algorithm described in Example 1.43 to compute the Hilbert function ofthe quotient rings R. Trivially, we already know that, for i < D, since the idealI has generators only in degree D,
HF(R; i) =
(n+ i
n
).
For the cases i = D + j with j ≥ 0, from Conjecture 4, we would have
HF(R; i) =∑
h<(k−1)(d−j)i−h∈kN
Nh
((n+ i−h
k
n
)−(n+ j
n
));
48
where Nh is simply the number of vectors h ∈ Zn+1k of weight wt(h) = h. In
order to compute the numbers Nh we may look at the following formula,
(k−1)(n+1)∑h=0
Nhxh = (1 + x+ . . .+ xk−1)n+1 =
(1− xk
1− x
)n+1
;
by expanding the right hand side, we get, for all h = 0, . . . , (k − 1)(n+ 1),
Nh =
bhk c∑s=0
(−1)s(n+ 1
s
)(n+ h− ks
n
).
Remark 1.48. From the conjectures, we would get also the extension of Corol-lary 1.42 in the k > 2 case, i.e.
[Rk,n,d]kd−1 = 0.
Indeed, with the same notation as above, let's take j = d−1. Thus, to computethe Hilbert function of the quotient in position kd− 1 we should compute theset Hd−1, i.e. the set of h ∈ Zn+1
k satisfying the following conditions:
kd− 1− wt(h) ∈ kZ, wt(h) < (k − 1)(d− d+ 1) = k − 1.
From the �rst condition, we get that wt(h) ∈ (k − 1) + kZ≥0 which is clearlyin contradiction with the second condition above. Thus, Hd−1 is empty andHF(R; kd− 1) = 0.
Example 1.49. Let's give one explicit example of the computations in orderto clarify the algorithm.
We consider the following parameters: k = 4, n = 2, d = 8. Thus we haveD = 24. Let's compute, for example, the Hilbert function of the correspondingquotient ring in degree i = 28, i.e. j = 4. Via the support of a computeralgebra software, as CoCoA5 by CoCoATeam [14] or Macaulay2 by Graysonand Stillman [28] and the implemented functions involving Gröbner basis, onecan see that
HF(R; 28) = 195.
Let's apply our algorithm to compute the same number. First, we need towrite down the vector N where, for l = 0 . . . (k − 1)(n + 1), Nl := #{h ∈Zn+1k | wt(h) = l}. In our numerical assumptions we have
N = (N0, . . . , N9) = (1, 3, 6, 10, 12, 12, 10, 6, 3, 1).
Now, we need to compute the vector H where we store all the possible weightsfor the vectors h ∈ H4, i.e. all the number 0 ≤ h ≤ 9 s.t. the following numericalconditions hold,
28− h ∈ 4Z, h < (k − 1)(d− j) = 12;
thus, H = (H0, H1, H2) = (0, 4, 8). Finally, we can compute HF(R; 28,h) foreach h ∈ H4. From our formula, it is clear that such numbers depend only on
49
the weight of h; thus, we just need to consider each single element in the vectorH. Assuming wt(h) = 0. We get,
R0 := HF(R; 28,0) = dimC S28,0 −(n+ j
n
)= 36− 15 = 21;
Similarly, we get: for wt(h) = 4,
R4 := HF(R; 28,h) = dimC S24,0 −(n+ j
n
)= 28− 15 = 13;
and, for wt(h) = 8,
R8 := HF(R; 28,h) = dimC S20,0 −(n+ j
n
)= 21− 15 = 6.
Now, we are able to compute the Hilbert function in degree 28.
HF(R; 28) = NH0RH0
+NH1RH1
+NH2RH2
=
= 21 + 12 · 13 + 3 · 6 = 21 + 156 + 18 = 195.
The algorithm.
In this section we want to show our algorithm implemented by using CoCoA5programming language, see CoCoATeam [14]. As we have seen in the previoussection, in the case k > 2, the algorithm is just conjectured. However, as wewill see in Section 1.60, we made several computer experiments supporting ourconjectures. Here is the CoCoA5 script of our algorithm based on Theorem 1.41and Remark 1.47.-- 1) Input parameters K, N, D;
K := ;
N := ;
D := ;
DD :=(K-1)*D;
-- HF will be the vector representing the Hilbert function
-- of the quotient ring;
HF :=[];
-- 2) Input vector NN where NN[I] counts the number of vectors
-- in ZZ^{n+1} modulo K of weight I;
Foreach H In 0..((N+1)*(K-1)) Do
M := 0;
Foreach S In 0..(Div(H,K)) Do
M := M+(-1)^S*Bin(N+1,S)*Bin(N+H-K*S,N);
EndForeach;
Append(Ref NN,M);
EndForeach;
50
-- 3) Compute the Hilbert Function:
-- in degree <DD:
Foreach L In 0..(DD-1) Do
Append(Ref HF,Bin(N+L,N));
EndForeach;
-- in degree =DD,..,K*D-1:
Foreach J In 0..(D-2) Do
I:=DD+J;
H:=[];
M:=0;
Foreach S In 0..I Do
If Mod(I-S,K)=0 Then
If S<(K-1)*(D-J) Then
If S<(K-1)*(N+1)+1 Then
Append(Ref H,S);
M:=M+1;
EndIf;
EndIf;
EndIf;
EndForeach;
HH:=0;
If M>0 Then
Foreach S In 1..M Do
HH:=HH+NN[H[S]+1]*(Bin(N+Div(I-H[S],K),N)-Bin(N+J,N));
EndForeach;
EndIf;
Append(Ref HF,HH);
EndForeach;
-- 4) Print the Hilbert function:
HF;
Remark 1.50. In the k = 2 case, our algorithm, which is proved to be true byTheorem 1.41, works very fast even with large values of n and d, e.g. n, d ∼ 300;cases that the computer algebra softwares, by involving the computation ofGröbner basis, cannot do in a reasonable amount of time and memory.
As regards the k > 2 case, with the support of computer algebra soft-ware Macaulay2 and its implemented function to compute Hilbert series ofquotient rings, we have checked that our numerical algorithm produces theright Hilbert function for two and three variables for low k and d. Moreover,in Section 1.3.3, we study the schemes of fat points related to our power ide-als and our results on their Hilbert series, will support Conjecture 4 in many
51
more cases. Precisely, with a combinatorial proof we check the n + 1 = 2case and, with the support of the computer algebra software CoCoA5, we havechecked that the conjectured algorithm gives the correct Hilbert function forall n, k ≤ 20 and d ≤ 150, see Remark 1.62.
1.3.3 Hilbert function of ξ-points in Pn
As we said in Section 1.1.3, we have a very close relation between power idealsand schemes of fat points; in particular, between their Hilbert functions becauseof Macaulay duality and Apolarity Lemma, see Theorem 1.18. In this Section,we want to study the schemes of fat points associated to the power ideals Ik,n,s.Precisely, we consider schemes of fat points with support over the kn points[1 : ξg1 : . . . : ξgn ] ∈ Pn, for any 0 ≤ gj ≤ k − 1. We call them ξ-points inshort. By using our results about the k = 2 case in Section 1.3.1, via Macaulayduality we get directly the Hilbert function for fat points with support on the(±1)-points. Consequently, after investigating the results for the k = 2 case,we have been able to compute the Hilbert function also for any k > 2.
The k = 2 case.
We begin by considering our class of power ideals in the k = 2 case, wherethe generators of the ideal Id are the d
th-powers of the 2n linear forms of typeL = x0 ± x1 ± . . .± xn. In Section 1.3.2, we have described an easy algorithmto compute the Hilbert function of the quotient rings Rd = S/Id, thus, viaMacaulay duality, we can apply our computations to get the Hilbert functionof schemes of fat points supported at all (±1)-points of Pn, namely the 2n
points of the type [1 : ±1 : . . . ± 1]. We'll see later that the results for thesearrangement of points can be directly extended to the k > 2 case.
Proposition 1.51. Let I(d) be the ideal associated to the scheme of d-fat pointssupported on the (±1)-points of Pn. Then,
HF(S/I(d),m) =
(n+mn
)for m ≤ 2d− 1(
n+2dn
)−(d+n−1n−1
)for m = 2d(
n+2d+1n
)− (n+ 1)
(d+n−1n−1
)for m = 2d+ 1
2n(n+d−1n
)for m ≥ 2d+ n− 2
Proof. By Corollary 1.42, we know that HF(Rm−d+1,m) = 0 for allm satisfyingthe inequality m ≥ 2(m− d+ 1)− 1 or, equivalently m ≤ 2d− 1; moreover, byRemark 1.45, we have that HF(Rd+1, 2d) =
(n+d−1n−1
), HF(Rd+2, 2d+ 1) = (n+
1)(n+d−1n−1
)and HF(Rm−d+1,m) =
(n+mn
)−2n
(n+d−1n
)for m ≤ 2(m−d+1)−n,
or equivalently, m ≥ 2d+ n− 2. By Macaulay duality, we are done.
Remark 1.52. Such result tell us that the ideal I(d) is generated in degree≥ 2d and, in particular, with
(d+n−1n−1
)generators in degree 2d. Thanks to the
geometrical meaning of the symbolic power I(d), we can easily �nd such gener-ators.
52
We may observe that we have exactly n pairs of hyperplanes which split our2n points. Namely, for any variable except xn, we can consider the hyperplanes
H+i = {xi + xn = 0} and H−i = {xi − xn = 0}, for all i = 0, . . . , n− 1.
It is clear that, for all i, half of our (±1)−points lie on H+i and half on H−i .
Consequently, we have n quadrics passing through our points exactly once, i.e.Qi = H+
i H−i = x2i − x2n, for all i = 0, . . . , n− 1.
Now, we want to �nd the generators of I(d), hence we want to �nd hyper-surfaces passing through our points with multiplicity d and we can consider,for example, all the monomials of degree d constructed with these quadricsQ0, . . . ,Qn−1, i.e. the degree 2d forms
G1 := Qd0, G2 := Qd−10 Q1, G3 := Qd−10 Q2, . . . ,GN := Qdn−1,
where N =(n+d−1n−1
). We can actually prove that they generate the part of
degree 2d of I(d) as a C-vector space. Since the number of Gi's is equal to thedimension of [I(d)]2d computed in Proposition 1.51, it is enough to prove thefollowing statement.
Claim. The Gi's are linearly independent over C.
Proof of the Claim. We prove it by double induction over the number of vari-ables n and the degree d. For two variables, i.e. n = 1, we have that thedimension of [I(d)]2d is equal to 1 for all d and then, G1 = Qd0 is the uniquegenerator. For n > 1, we consider �rst the d = 1 case. Assume to have a linearcombination
α0Q0 + . . .+ αn−1Qn−1 = α0(x20 − x2n) + . . .+ αn−1(x2n−1 − x2n) = 0.
Specializing on the hyperplane H−0 = {x0 = xn}, we reduce the linear combina-tion in one variable less and, by induction, we have αi = 0 for all i = 1, . . . , n−1;consequently, also α0 = 0.
Assume to have a linear combination for d ≥ 2, namely
α1G1 + α2G2 + . . .+ αNGN =
= α1(x20 − x2n)d + α2(x20 − x2n)d−1(x21 − x2n) + . . .+ αN (x20 − x2n)d = 0.
By specializing again on the hyperplane H−0 = {x0 = xn}, we get a linearcombination in the same degree but with one variable less and, by inductionover n, we have that αi = 0 for all i where the de�nition Gi doesn't involve(x20 − x2n)d. Thus, we remain with a linear combination of type
(x20 − x2n)[α0Qd−10 + α1Qd−20 Q1 + . . .+ αmQd−1n−1
]= 0;
by induction over d, we are done.
53
Hence, we can consider the ideal Jd = (x20 − x2n, . . . , x2n−1 − x2n)d. It is clearly
contained in I(d) but, a priori, it could be smaller. In order to show that theequality holds and that I(d) is minimally generated by the Gi's, we start bystudying the Hilbert series of the ideal Jd.
Lemma 1.53. Let Td = C[x0, . . . , xn]/Jd, where Jd = (x20 − x2n, . . . , x2n−1 −x2n)d, then the Hilbert series is
HS(Td; t) =1 +
∑ni=1(−1)iβit
2d+2(i−1)
(1− t)n+1,
where βi := βi,2d+2(i−1) =(d+i−2i−1
)(d+n−1n−i
), for all i = 1, . . . , n, and the multi-
plicity is e(Td) = 2n(d+n−1n
).
Proof. The quotient Td is a 1-dimensional Cohen-Macaulay ring and xn is anon-zero divisor. Thus, we have that Td and the quotient Td/(xn) have thesame Betti numbers; moreover, we have that
Td/(xn) = C[x0, . . . , xn−1]/(x20, . . . , x2n−1)d,
and the resolution of those quotients are very well known. The quotient ringC[x0, . . . , xn]/(x0, . . . , xn−1)d has a pure resolution of type (d, d+1, . . . , d+n−1)and its Betti numbers and multiplicity are expressed with an explicit formula,see Theorem 4.1.15 in Bruns and Herzog [10].
Thus, Td/(xn) has a pure resolution of type (2d, 2d+2, 2d+4, . . . , 2d+2(n−1)), i.e.
. . . −→ S(−2d− 4)β3,2d+4 −→ S(−2d− 2)β2,2d+2 −→ S(−2d)β1,2d −→ 0,
where S is the graded polynomial ring C[x0, . . . , xn−1] and S(−i) is its ith-shifting, i.e. [S(−i)]j := Sj−i. Moreover, the Betti numbers and the multiplicityof the quotient are given by the following formulas,
βi := βi,2d+2(i−1) = (−1)i+1∏j 6=i
d+ j − 1
j − i=
=����
(−1)i+1 d(d+ 1) · · · (d+ i− 2)
����(−1)i−1(i− 1)!
· (d+ i) · · · (d+ n− 1)
(n− i)!=
=
(d+ i− 2
i− 1
)(d+ n− 1
n− i
);
e(Td) =1
n!
n∏i=1
(2d+ 2(i− 1)) = 2n(d+ n− 1
n
).
From the Betti numbers, we can easily get the Hilbert series of Td = S/Jd,
HS(Td; t) =1 +
∑ni=1(−1)iβit
2d+2(i−1)
(1− t)n+1.
54
Corollary 1.54. Let Td = C[x0, . . . , xn]/Jd, where Jd = (x20 − x2n, . . . , x2n−1 −x2n)d, then
HF(Td,m) =
(n+mn
)for m ≤ 2d− 1(
n+2dn
)−(d+n−1n−1
)for m = 2d(
n+2d+1n
)− (n+ 1)
(d+n−1n−1
)for m = 2d+ 1
2n(n+d−1n
)for m� 0
Proof. The values of the Hilbert function for m ≤ 2d + 1 follow directly byextending the Hilbert series computed in Lemma 1.53, recalling that 1
(1−t)n+1 =∑i≥0(n+in
)ti. Moreover, since Td is a 1-dimensional CM ring, we have that its
Hilbert function is eventually constant and equal to the multiplicity.
Now, we are able to complete our study of the ideal of fat points with supporton the (±1)-points in Pn.
Theorem 1.55. Let I(d) be the ideal associated to the scheme of fat points
of multiplicity d and support on the 2n points [1 : ±1 : . . . : ±1] ∈ Pn. Thegenerators are given by the monomials of degree d made with the n quadrics
Qi = x2i − x2n, for all i = 0, . . . , n− 1, and the Hilbert series is
HS(S/I(d); t
)=
1 +∑ni=1(−1)iβit
2d+2(i−1)
(1− t)n+1,
where the Betti numbers are given by
βi := βi,2d+2(i−1) =
(d+ i− 2
i− 1
)(d+ n− 1
n− i
), for i = 1, . . . , n.
Proof. Let's write I(d) = Jd + J where Jd = (Q0, . . . ,Qn−1)d. From Lemma1.53, it is enough to show that J = 0. We consider the quotient Td = S/(I(d) +(xn)) = C[x0, . . . , xn−1]/((x20, . . . , xn)d + J̄) and the exact sequence
0 −→ Ann(xn) −→ S/I(d)·xn−→ S/I(d) −→ Td −→ 0.
Consequently, we get
HS(Td; t) = (1− t) HS(S/I(t); t) + HS(Ann(xn); t).
Since S/I(d) is 1-dimensional ring, we have that HS(S/I(t); t) = h(t)(1−t) and the
multiplicity is given by e(S/I(d)) = h(1). Thus, the multiplicity of Td is givenby
e(Td) = h(1) + HS(Ann(xn); 1) ≥ e(S/I(d)) = 2n(d+ n− 1
n
); (1.6)
moreover, the equality holds if and only if xn is a non-zerodivisor of Td. Onthe other hand, we have that Td = C[x0, . . . , xn−1]/(x20, . . . , xn−1)d + J̄ andconsequently, by Lemma 1.53, we have
e(Td) ≤ e(C[x0, . . . , xn−1]/(x20, . . . , xn−1)d
)= 2n
(d+ n− 1
n
); (1.7)
55
where equality holds if and only if J̄ = 0. From (1.6) and (1.7), we can concludethat
� xn is a non-zerodivisor for Td = S/I(d);
� J̄ = 0.
Now, let's assume J 6= 0 and take a non-zero element f ∈ J of minimal degreein J . Then, since J̄ = 0, we get that f = xn · g, for some g, thus we havexn ·g = 0 in Td. This contradicts that xn is a non-zerodivisor in Td, since g /∈ Jbecause of minimality of f in J and g /∈ Jd because f is not.
Remark 1.56. In the last decades, the study of the behavior between symbolic
and regular powers of homogeneous ideals involved many mathematicians anddi�erent areas. By de�nition, we always have the inclusion Im ⊂ I(m), but theequality is not always true. Consequently, people started to study containment
problems, as in Ein, Lazarsfeld, and Smith [17] and Hochster and Huneke [30].In Bruns and Herzog [10], the author showed that for any c < n, there existsan ideal of points in Pn such that I(m) * Ir for some m > cr. In Bocci,Cooper, and Harbourne [8], there is a list of open conjectures regarding thiscontainment problems. The authors showed also that all the conjectures holdin case of equality between symbolic and regular powers I(m) = Im for any m.
Our ideals of points in Pn satisfy always the equality between symbolic andregular powers; consequently, they satisfy all the conjectures listed in Bocci,Cooper, and Harbourne [8].
Even from the point of view of Gröbner basis, our result is very useful. If we �xan ordering on the variables, a Gröbner basis for the ideal I is simply a set ofgenerators such that their initial terms generate the initial ideal in(I); see e.g.Cox, Little, and O'Shea [16]. We recollect such properties in the following.
Corollary 1.57. Let I(d) be the ideal of fat points of multiplicity d supported onthe (±1)-points of Pn. Then, we have the equality between I(d) = Id. Moreover,
for any ordering such that xn > xi for all i = 0, . . . , n−1, the set of generatorsgiven in Theorem 1.55 is actually a Gröbner basis for I(d).
Proof. It follows directly from Theorem 1.55, since we have thatI = I(1) = (x20 − x2k, . . . , x2n−1 − x2n).
Moreover, considering the Gi's, i.e. the set of generators obtained by takingall the possible monomial of degree d in the quadrics xi − xn, for all i =0, . . . , n−1, we have that their leading terms generate the initial ideal, i.e. theyare a Gröbner basis. Indeed, we clearly have the inclusion
(in(Gi)) ⊂ in(I);
but, we also have that the left hand side is exactly (in(Gi)) = (x20, . . . , x2n−1)d,
which has the same Hilbert function of I, as we have seen in the proof ofTheorem 1.55, and consequently the same Hilbert function of in(I). Hence, theequality holds.
56
The k > 2 case.
Let ξ be a kth-root of unity and consider the ideal I(d)k corresponding to the
scheme of fat points of multiplicity d and support on the kn ξ-points of type[1 : ξg1 : . . . : ξgn ] ∈ Pn with 0 ≤ gi ≤ k − 1, for all i = 1, . . . , n.
In Section 1.3.2, we have considered the power ideals In,k,d related to suchpoints where the powers where only multiples of (k−1). Thus, we cannot hopeto get the Hilbert series of our scheme of fat points directly from our previousresults on the Hilbert series of Rn,k,d = S/In,k,d. However, we can easily observethe following,
HF(I(d), kd− 1
)= HF (Rn,k,d, kd− 1) ;
from Remark 1.48, we get that, assuming true the Hilbert function of Rk,d
conjectured, the ideal I(d)k should be generated at least in degree kd. Thus,
inspired by the k = 2 case, we can actually claim that I(d)k is nonzero in degree
kd. Indeed, we have that, for any variable x0, . . . , xn−1, we can consider the khyperplanes
H0i = {xi − xn = 0}, H1
i = {xi − ξxn = 0}, . . . , Hk−1i = {xi − ξk−1xn = 0};
such hyperplanes divide the kn points in k distinct groups of kn−1 points; thus,their products give a set of degree k forms which vanish with multiplicity 1 ateach point, i.e.
Qi = H0i ·H1
i · · ·Hk−1i = xki − xkn, for all i = 0, . . . , n− 1.
Consequently, we get
Jk,d = (Q0,Q1, . . . ,Qn−1)d ⊂ I(d)k .
Now, by using the same ideas as for the k = 2 case, we can get the analogousof Lemma 1.53 and Theorem 1.55 for all k ≥ 2 and consequently we get thefollowing general result.
Theorem 1.58. Let I(d)k be the ideal associated to the scheme of fat points
of multiplicity d and support on the kn ξ-points [1 : ξg1 : . . . : ξgn ] ∈ Pn for
0 ≤ gi ≤ k − 1. The generators are given by the monomials of degree d made
with the n forms of degree k Qi = xki − xkn, for all i = 0, . . . , n − 1 and the
Hilbert series is
HS(S/I
(d)k ; t
)=
1 +∑ni=1(−1)iβit
kd+k(i−1)
(1− t)n+1,
where the Betti numbers are given by
βi := βi,kd+k(i−1) =
(d+ i− 2
i− 1
)(d+ n− 1
n− i
), for i = 1, . . . , n.
Remark 1.59. Moreover, similarly as for Corollary 1.57, we have that
57
� I(d)k = Idk ;
� the set of generators given in the theorem above, is a Gröbner basis.
Remark 1.60. Since we have explicitly computed the Hilbert series of ξ-pointsin Pn, by using again Macaulay duality, we can go back to look at the Hilbertseries of the power ideals In,k,d. In particular, we can check that our Conjecture4 holds in a lot of cases.
Let Rn,k,d be the quotient ring S/In,k,d where In,k,d is the power idealgenerated by all the (x0 + ξg1x1 + . . . + ξgnxn)(k−1)d with 0 ≤ gi ≤ k − 1 for
all i = 1, . . . , n; and let I(d)k be the ideal associated to the scheme of fat points
of multiplicity d and support on the ξ-points of Pn.Now, we have seen in Section 1.3.2 that, since In,k,d is generated in degree
(k − 1)d and generate the whole space in degree kd − 1, the Hilbert functionof Rn,k,d has to be computed only in the degrees i = (k − 1)d + j, with j =0, . . . , d− 2. In that degrees, by Macaulay duality, we get
HF(Rn,k,d; i) = HF(I(j+1)k ; i
).
From Theorem 1.58, we can explicitly compute such Hilbert function, i.e.for all j = 0, . . . , d− 2,
HF(Rn,k,d; i) = (1.8)
=∑s∈N
s≤ k−1k (d−j)
(−1)s+1
(n+ (k − 1)(d− j)− ks
n
)(j + s− 1
s− 1
)(j + n
n− s
).
In Section 1.3.2, we conjectured an extension of our formula for the Hilbertseries of the quotient Rn,k,d based on a Zn+1
k -grading on the polynomial ring.We may recall the formula conjectured: for all j = 0, . . . , d− 2,
HF(Rn,k,d; i) =∑
h<(k−1)(d−j)i−h∈kN
Nh
((n+ i−h
k
n
)−(n+ j
n
)); (1.9)
where Nh is simply the number of vectors h ∈ Zn+1k of weight wt(h) = h, see
Remark 1.47. In order to show that formula (1.9) is right and then to proveConjecture 4, we should show that the right hand side of such formula is equalto the right hand side of formula (1.8).
Proposition 1.61. Assuming n = 1, i.e. in the two variables case, the formu-
las (1.8) and (1.9) are equal and Conjecture 4 is true.
Proof. For any k and d, the unique non-zero addend is the one for s = 1; thus,
(1.8) = 1 + (k − 1)(d− j)− k.
58
Now, we look at formula (1.9). First of all we may observe that, for n = 1,the number of vectors in Z2
k with �xed weight h can be computed very easily,indeed
Nh =
{h+ 1 for 0 ≤ h ≤ k − 1;
2k − (h+ 1) for k ≤ h ≤ 2(k − 1).
Thus, any i = (k − 1)d+ j can be written as ck + r for some positive integersc, r with 0 ≤ r ≤ k − 1 and then, we get
(1.9) = Nr(1 + c− (j + 1)) +Nr+k(1 + (c− 1)− (j + 1)) =
= (r + 1)(1 + c− (j + 1)) + (k − r − 1)(1 + (c− 1)− (j + 1)) =
=����(r + 1)c+ r + 1−((((
(((r + 1)(j + 1) + kc−����(r + 1)c− kj − k +(((((((r + 1)(c+ 1);
moreover, recalling that i = ck + r = (k − 1)d+ j, we �nally get
(1.9) = 1 + (k − 1)d+ j − kj − k = 1 + (k − 1)(d− j)− k.
Remark 1.62. With similar, but longer and more intricate arguments as forProposition 1.3.3, we have been able to check also the case n + 1 = 3. Un-fortunately, we have not been able to prove that the two expressions given in(1.8) and (1.9) give the same answer for any possible parameters (k, n, d). Withthe support of a computer, by implementing with the CoCoA5 language suchformulas, we have been able to check all the cases n, k ≤ 20, d ≤ 150.
Here is the implementation of the formula (1.8) by using CoCoA5 language. Asregards the formula (1.9), we have used the algorithm in Section 1.3.2.
-- 1) Input of the parameters K, N, D;
K := ;
N := ;
D := ;
DD := (K-1)*D;
-- HF will be the vector containing the relevant part of
-- the Hilbert function, i.e. from (K-1)D to KD-2;
HF := [];
-- 2) Compute the Hilbert function;
Foreach J In 0..(D-2) Do
B := 0;
KK := (K-1)*(D-J)/K;
Foreach S In 1..N Do
If S <= KK Then
B := B+(-1)^(S+1)*Bin(N+(K-1)*(D-J)-K*S,N)*
Bin(J+S-1,S-1)*Bin(J+N,N-S);
EndIf;
EndForeach;
59
Append(Ref HF , B );
EndForeach;
-- 3) Print the Hilbert function;
HF;
60
CHAPTER 2
The Waring problem for polynomials
Section 2.1
Introduction
2.1.1 Historical background 1
In 1770 the number theorist Edward Waring, in his Meditationes Algebraicae,stated the following
every natural number can be written as sum of at most 4 squares;
every natural number can be written as sum of at most 9 cubes;
every natural number can be written as sum of at most 19 fourth powers;
and so on...
Apparently, he had in my the following general statement,
for any number d ≤ 2, there exists a minimal number g(d) such that
every natural number can be written as sum of at most g(d) dth-power.
Lagrange's theorem gives us that g(2) = 4; however, to have a proof of theexistence of such a number g(d) for any d ≥ 2, we had to wait until 1909 whenDavid Hilbert gave the proof of the theorem. However, even if g(3) = 9 wasknown and g(4) = 19 has been determined a few years ago, in general the valueof g(d) is unknown. There are bounds and equations that this number satis�es,but we don't want to go into the details of this numerical problem.
Furthermore, it has been observed that, even if g(3) = 9, only �nitely manynatural numbers require either 8 or 9 cubes; for any integer su�ciently large,only 7 cubes are required. Hence, for any number d ≥ 2, we can also de�ne thenumber G(d) as the minimal number required for any su�ciently large naturalnumber to being written as sum of dth-powers.
1See Geramita [26] for a deeper survey on the problems and their connections.
61
By de�nition, we have G(d) ≤ g(d), but, as we said, this inequality can bestrict and for example G(2) = g(2) = 4, G(3) = 7 and G(4) = 16. In general,very little is known about the number G(d).
We are interested in the polynomial version of that problem.
Let F be a homogeneous polynomial of degree d in a polynomial ring S.Can we write F as sum of dth-powers of linear forms?
The answer is immediately positive since, by Macaulay duality, we have thatthe dimension of the vector space generated by s generic dth-powers of linearforms is equal to the Hilbert function of s generic simple points computed indegree d. Now, this Hilbert function is equal to the minimum between s andthe dimension of the ambient space Sd; hence, for s su�ciently large we havethat s generic dth-powers of linear forms generated the whole space Sd.
However, we can ask what is the minimal way to write such a form Fas sum of powers and, furthermore, what is the minimal number of linearforms required by any form of degree d. Hence, the equivalent of the numericalquestion regarding the computation of g(d) is given by the following.
�Small� Waring problem: �xed a number d ≥ 2, what is the minimalinteger r such that any form of degree d can be written as sum of at mostg(d) dth-powers of linear forms? Such minimal number is called Waring rank,or simply rank, of F . We denote it as #d(F ).
Moreover, as for the numerical case, we can try to ask the same question byconsidering almost all the forms of degree d. In other words, similarly to considerthe numerical Waring problem only for su�ciently large numbers, in terms ofpolynomials we can look at Zariski open (dense) subsets of the whole space Sd.
�Big� Waring problem: �xed a number d ≥ 2, what is the Waring rankof the generic form of degree d?
As we are going to see later, the Waring problem for polynomials can be in-terpreted in terms of secant varieties of Veronese varieties or in terms of fatpoints. Because of those relations, the answer for the �Big� problem is given bythe Alexander-Hirschowitz Theorem 1.11.
Instead, the �Small� Waring is largely unknown. The answer is known fortwo variables and for monomials in any number of variables. In Comas andSeiguer [15], there is described an algorithm which computes the rank of anybinary polynomial given in input and, with some exceptions, provides also thelinear forms required. In Carlini, Catalisano, and Geramita [11], the authorsprovide a formula which computes the rank of a monomial. We see this formulalater, in Section 2.4.
Remark 2.1. Due to clear historical reasons, in the bibliography, all problemsabout some additive decompositions are called Waring problems. In particu-lar, the Waring problem for tensors has received a lot of attention in the lastdecades from mathematicians, but also from engineers because of its practicalapplications. The idea is brie�y the following.
Let V1, . . . , Vn be vector spaces over the same �eld. We may consider theirtensor product V1⊗ . . .⊗Vn. Inside this vector space we have the decomposable
62
tensors of the type v1⊗. . .⊗vn. The Waring problem for tensors is the following.
Given a tensor T ∈ V1 ⊗ . . .⊗ Vn, what is the minimal number of
decomposable tensors required to write T as their sum?
What about the generic tensor?
For tensors, not even the generic case has been solved in all generality. Due tothe geometrical interpretation in terms of secant varieties of Segre varieties, weknow the answer for generic tensors of the following types:
• products of two vector spaces; a very classical result, e.g. see Landsberg[32, Proposition 5.3.1.4];
• products of copies of vector spaces of dimension 2, see Catalisano, Geramita,and Gimigliano [13];
• unbalanced products, namely when the dimension of one vector space ismuch bigger than the others, see Catalisano, Geramita, and Gimigliano [12].
Otherwise, there are only partial results on secant varieties of Segre varieties inparticular cases, e.g. in Abo, Ottaviani, and Peterson [1]. For a complete list ofthese results and a description of the geometry of tensors and their applicationssee Landsberg [32].
2.1.2 The geometry of the problem
We want to bree�y explain the geometry behind the Waring problem for poly-nomials. First of all, we observe that the objects which describe powers of linearforms are the Veronese varieties.
Veronese varieties and their secant varieties
De�nition 2.2. The Veronese variety Vn,d is the image of the regular map
νd : Pn −→ Pd
[a0 : . . . : an] 7−→ [ad0 : ad−10 a1 : . . . : adn].
Let S = C[x0, . . . , xn] with the standard gradation S =⊕
i∈N Si. By consid-ering as basis of the C-vector spaces Sd the usual monomial basis modi�ed
by
mi0,...,in :=
(d
i0, . . . , in
)xi00 . . . x
inn , for all i0 + . . .+ in = d, with all is ≥ 0;
we may observe that the Veronese embedding can be restated as follows,
νn,d : P(S1) −→ P(Sd),
[L] 7−→ [Ld].
Hence, we can say that
63
Veronese variety Vn,d parametrizes the dth-powers of linear forms
in n+ 1 variables inside the space of forms of degree d.
Now, let's take two distinct points on the Veronese variety Vn,d; namely, [Ld]and [Md], where L,M ∈ S1. Then, consider the line spanned by those twopoints l. By de�nition, any point [F ] ∈ P(Sd) which lies on that line l is alinear combination of the two points spanning the line. In other words, we canwrite [F ] = [Ld] + [Md] and, in terms of Waring problem, the homogeneouspolynomial F has Waring rank 2. More in general, the degree d form corre-sponding to a point [F ] ∈ P(Sd) which lies on a (s − 1)-dimensional linearspace spanned by s points on our Veronese variety Vn,d has Waring rank s.
[L ] [L ]
[L ][L ]+[L ]+[L ]
1 2
31 2 3
Vn,d
dd
d
d d d
Figure 2.1: Generic points on a secant plane spanned by three independentpoints of Veronese variety have Waring rank equal to 3.
Hence, the geometrical objects we need to look at are the secant varieties toVeronese varieties.
Given any projective variety X ⊂ Pn, we can de�ne its �rst secant varietyby considering the union of all possible lines passing through two distinct pointsof the variety X and then taking the closure in the Zariski topology, in orderto get a variety. Geometrically, by taking the Zariski closure, we are simplyconsidering also the tangent lines.
In general, we de�ne the sth-secant variety of X as follows.
De�nition 2.3. Let X be a projective variety. The sth-secant variety σs(X)is given by taking the Zariski closure of the union of all Ps−1's spanned by
64
independent points of X, i.e.
σs(X) =⋃
P1,...,Ps∈Xindependents
〈P1, . . . , Ps〉Zariski
⊂ Pn.
Coming back to our �Big� Waring problem, we may observe that, in order tocompute the Waring rank of the generic form of degree d, it is enough to com-pute the smallest secant variety σs(Vn,d) which �lls the ambient space P(Sd).The existence of such a secant variety follows from the fact that Veronese vari-eties are non-degenerate, namely they are not contained in a proper linear sub-space of the ambient space. Hence, one can prove that: given a non-degeneratevariety X ⊂ Pn, we have a strict chain of inclusions
X ( σ2(X) ( . . . ( σs(X) = Pn.
In this way, the �Big� Waring problem becomes a question about the dimensionof secant varieties of Veronese varieties. By a simple count of parameters, wecan de�ne the expected dimension of such varieties; namely,
exp dimσs(Vn,d) = min{sn+ s− 1, N}, where N =
(n+ d
n
)− 1.
Unfortunately, the expected dimension is not always the correct one and it isvery easy to �nd a Veronese variety which has a secant variety with dimensionsmaller than expected.
De�nition 2.4. If the sth-secant variety of a projective variety X has dimen-sion smaller than the expected, we say the X is s-defective.
Example 2.5. Veronese surface in P5 is 2-defective. Let S = C[x, y, z].We consider the space of plane conics P(S2) and the Veronese surface V2,2contained inside such P5. Any plane conic can be expressed as a symmetric3 × 3 matrix. Hence, the Veronese variety V2,2 can be seen as the space ofsymmetric 3 × 3 matrices of rank 1. The expected dimension of σ2(V2,2) is 5and then we expect that it �lls the whole ambient space.
However, a linear combination of two rank 1 matrices can have at most rank2; consequently, we have that σ2(V2,2) is contained in the hypersurface of P5
de�ned by the determinant. In particular, the dimension is at most 4. 1
We have seen how our problem became a question about the dimension of thesecant varieties of Veronese varieties. A way to compute the dimension of avariety is to look at its tangent space TP (X) in a generic point: the dimensionof the variety will be equal to the dimension of this tangent space. More often,we look at the a�ne dimension of the underlying vector space of this tangentspace which, in order to avoid confusion, we denote as TP (X). The dimensionof the projective variety X is one less than the a�ne dimension of TP (X).
1Actually, σ2(V2,2) is the precisely hypersurface de�ned by the determinant.
65
Theorem 2.6 (Terracini's Lemma [33]). Let X be a projective variety and
P1, . . . , Ps be points of X in generic position. Let P a generic point over the
span of the points P1, . . . , Ps; then, the tangent space of σs(X) at the point Pis equal to the span of the tangent spaces of X at the points Pi's; namely
TP (σs(X)) = span{TP1(X), . . . , TPs
(X)}.
Terracini's Lemma gives us a very direct algorithm to compute the dimensionof secant varieties σs(Vn,d): compute the tangent space of Vn,d and use theprojective Grassmann's formula to compute the dimension of the span of stangent spaces of Vn,d at s generic points.
Example 2.7. Yet again: Veronese surface in P5 is defective. Let'sconsider again the Veronese surface in P5. By Terracini's Lemma, we have thatthe tangent space of the second secant variety is spanned by two tangent spacesin two di�erent points of the Veronese surface. Hence, by Grassmann projectiveformula, since in general two planes in P5 do not intersect, we expect that thedimension of the tangent space of our secant variety is equal to 5 1.
However, it is easy to see that, the intersection of two tangent spaces ofthe Veronese surface V2,2 in two distinct points [L2] and [M2], intersect in onepoint; namely,
T[L2](V2,2) ∩ T[M2](V2,2) = [LM ].
Thus, the dimension of the tangent space is actually 4 and we get that V2,2 is2-defective.
As we have just seen in the previous example, Terracini's Lemma allows us tocompute the dimension of the tangent spaces of secant varieties and then, apriori, to �nd the solution for the �Big� Waring problem. However, we cannotuse it directly to solve the problem in its full generality, for any possible d andn.
The idea is to use Terracini's Lemma to change once again the setup of ourquestion and to connect the Waring problem in terms of Hilbert series of powerideals and fat points.
Let L1, . . . , Ls be generic linear forms and let Id = (Ld1, . . . , Lds). Then, by
Terracini's Lemma
dimσs(Vn,d) = dim(Ld−11 S1 + . . .+ Ld−1s S1) = dim[Id−1]d;
now, consider the points Pi = [Li] ∈ Pn and their corresponding prime ideals℘i, for all i = 1, . . . , s, respectively, and let I(2) = ℘2
1 ∩ . . . ∩ ℘2s be the ideal
of the scheme of double points with support at the Pi's. By Apolarity Lemma1.18, we have that
dim[Id−1]d = dim[S/I(2)]d.
In conclusion, we have reduced the �Big� Waring problem �rst to a questionabout dimension of secant varieties of the Veronese varieties and then to aquestion about the Hilbert function of double points. As we have seen in Section
1Recall that, by de�nition, the dimension of the empty space is −1.
66
[LM]
[M ][L ]2 2
V ⊂ℙ2,25
Figure 2.2: The Veronese surface V2,2 in P5 is 2-defective since two generic tan-gent spaces, although we expect they have empty intersection, they do intersectin one single point.
1.1.3, this last question has been completely solved by Alexander-HirschowitzTheorem 1.11 and we get, as direct corollaries, the following solutions for thequestion about dimension of secant varieties and for the �Big� Waring problem.
Corollary 2.8. Let F be a generic form of degree d in n+ 1 variables. Then,
#d(F ) =
⌈(d+nn
)n+ 1
⌉;
except for the following exceptions,
1. d = 2 where #2(F ) = n+ 1;
2. d = 4 and n = 2 where #4(F ) = 6;
3. d = 4 and n = 3 where #4(F ) = 10;
4. d = 3 and n = 4 where #3(F ) = 8;
5. d = 4 and n = 4 where #4(F ) = 15.
Corollary 2.9. The Veronese varieties are non-defective, namely
dimσs(Vn,d) = min
{sn+ s− 1,
(n+ d
n
)};
except for the following exceptions,
67
1. d = 2 and 2 ≤ s ≤ n with defect equal to(n−s+2
2
);
2. n = 2, d = 4 and s = 5 with defect 1;
3. n = 3, d = 4 and s = 9 with defect 1;
4. n = 4, d = 3, 4 and s = 7, 14, respectively; both with defect 1.
Section 2.2
Sum of squares
2.2.1 A generalized Waring problem
In Fröberg, Ottaviani, and Shapiro [25], the authors began the study of ageneralized Waring problem.
Let S = C[x0, . . . , xn] be the polynomial ring in n+ 1 variables.
Generalized Waring problem: let k and d be positive integers withk ≥ 2. Given a generic form F ∈ Skd, what is the minimal number of degree dforms such that F can be written as sum of their kth-powers? We call such anumber kth-Waring rank and we denote it as #k(d, n).
By a simple count of parameters, we can guess what is the solution of ourquestion for each triple (k, d, n), i.e.
exp#k(d, n) =
⌈dim Skdndim Sdn
⌉.
Remark 2.10. The d = 1 case is the original �Big� Waring problem solved byJ. Alexander and A. Hirschowitz, see Corollary 2.8.
For d ≥ 2, we do not have a complete solution. In Fröberg, Ottaviani,and Shapiro [25], the main result is an upper-bound for the kth-Waring rank#k(d, n) of a generic form. Moreover, this upper bound does not depend on dand, asymptotically in d, it is also sharp.
Theorem 2.11. (Fröberg, Ottaviani, and Shapiro [25, Theorem 4])Let k, d and n be as above. A generic form of degree kd in n+ 1 variables can
be written as sum of at most kn kth-powers of forms of degree d.
From this result we can easily get the solution of the generalized Waring prob-lem in the case of binary forms of even degree, namely #2(d, 1) = 2. However,with an easy algebraic argumentation, we can see that actually all the binaryforms of even degree have 2nd-Waring rank equal to 2.
Theorem 2.12. (Fröberg, Ottaviani, and Shapiro [25, Theorem 5])Any form of degree 2d in two variables can be written as sum of at most 2squares.
68
Proof. For any F of degree 2d in two variables, there exist two degree d formsG and H such that we can write
F = G ·H =
(1
2(G+H)
)2
+
(i
2(G−H)
)2
.
Similarly to the original Waring problem, we can give a geometric interpretationto this generalized question.
Given k, d and n as above, we can consider the variety of kth-powers
Vn,k,d := {[Gk] | G ∈ Sd} ⊂ P(Skd).
Such variety has dimension equal to(n+dn
)−1 and the tangent space at a point
[Gk] is given by
T[Gk] ={
[Gk−1H] | H ∈ Sd}
;
consequently, by Terracini's Lemma, we have that the sth-secant varietyσs(Vn,k,d) has dimension one less than the dimension of the homogeneous partof degree kd of the ideal (Gk−11 , . . . , Gk−1s ), where G1, . . . , Gs are generic formsof degree d.
Theorem 2.13. Fröberg, Ottaviani, and Shapiro [25, Theorem 6] Thegeneric form of degree kd can written as sum of Gk1 , . . . , G
ks , where all the Gi's
have degree d, if and only if the ideal (Gk−11 , . . . , Gk−1s ) contains all the space
Skd.
Remark 2.14. Actually, in Fröberg, Ottaviani, and Shapiro [25], the authorsproved Theorem 2.11 as a consequence of Theorem 2.13. Indeed, they consid-ered the power ideals we considered in Section 1.3, namely
In,k,d =(
(x0 + ξg1x1 + . . .+ ξgnxn)(k−1)d∣∣ ∀i = 1, . . . , n, 1 ≤ gi ≤ k − 1
).
They proved that the ideal In,k,d contains the space Skd. Clearly, such result canonly give an upper bound for the solution since they considered a very specialset of forms of degree d. Moreover, the result is even more surprising since thesegenerators are already powers of linear forms. Geometrically speaking, we aretaking our points in a small subvariety of dimension n, namely the subvarietycoming from the Veronese variety of dth-powers of linear forms in n+1 variableswhich is contained in our variety Vn,k,d which has dimension N =
(n+dn
)− 1,
and this choice give asymptotically the best answer.
Remark 2.15. By Theorem 2.13, we have related the generalized Waringproblem again to a question about some sort of power ideals. Unfortunately,for ideals generated by powers of forms of degree higher than 1, we do nothave an analogue of the Apolarity Lemma 1.18 and we cannot relate our newquestion to the computation of Hilbert functions of schemes in Pn.
69
In this section we consider the generalized Waring problem in the case ofk = 2; namely, we consider a generic form of even degree 2d and we want towrite this form as a sum of squares.
As we have seen in Theorem 2.12, the k = 2 case in two variables can besolved very easily and we have that #2(d, 1) = 2. In general, from Theorem2.11, we only have the upper-bound #2(d, n) ≤ 2n which is eventually sharp.Our aim is to consider the case of three and four variables, namely n = 2, 3respectively. In those cases, we complete the graph of the function #2(d, n) alsofor low d, before reaching the asymptotic value 2n.
2.2.2 Sum of squares in three variables.
By Theorem 2.11, we have that
#2(d, 2) ≤ 4.
By a simple count of parameters, we have that
exp#2(d, 2) =
⌈dimS2d
dimSd
⌉=
⌈(2d+ 2)(2d+ 1)
(d+ 2)(d+ 1)
⌉=
⌈2(2d+ 1)
(d+ 2)
⌉. (2.1)
Since the expected 2nd-Waring rank is always a lower bound for the exact an-swer, we get #2(d, 2) = 4, for d large enough. More precisely, for all d > 4.Indeed, ⌈
2(2d+ 1)
(d+ 2)
⌉> 3 ⇐⇒ 4d+ 2 > 3d+ 6 ⇐⇒ d > 4.
Thus, in order to complete the answer in the three variables case, we just needto check what happen for d ≤ 4.
For d = 1, we just go back to the Alexander-Hirschowitz result, see Corollary2.8, which gives
#2(1, 2) = 3.
For d = 2, we claim that #2(2, 2) = 3. By the formula (2.1), we have that3 ≤ #2(2, 2) ≤ 4. Hence, by Theorem 2.13, it is enough to show that
given three generic forms of degree 2 in three variables G1, G2, G3,the ideal I = (G1, G2, G3) contains in degree 4 the whole space S4.
From an algebraic point of view, we want to look at the Hilbert function of thequotient ring C[x0, x1, x3]/(G1, G2, G3). If we can show that in degree 4, suchHilbert function is 0, we are done.
Since we have three generic forms in three variables, we have that I is a completeintersection and then, as we saw in Example 1.5, we have that the Hilbert seriesof S/I is equal to
i 0 1 2 3 4HF(S/I; i) 1 3 3 1 -
70
In particular, HF(S/I; 4) = 0 and then #2(2, 2) = 3.
Again, in the d = 3 case, we claim #2(3, 2) = 3. By formula (2.1), we havethat 3 ≤ #2(2, 2) ≤ 4 and again it is enough to show that a generic idealI = (G1, G2, G3) generated by three forms of degree 3 contains the whole spaceS6. Since we are again looking at an ideal which is a complete intersection, wehave that the Hilbert function of S/I is given by
i 0 1 2 3 4 5 6 7HF(S/I; i) 1 3 6 7 6 3 1 -
In particular, HF(S/I; 6) 6= 0 and the #2(3, 2) = 4.
We proceed similarly for d = 4. By formula (2.1), we have that 3 ≤#2(4, 2) ≤ 4. We look at the Hilbert function of S/I with I = (G1, G2, G3)a generic ideal generated by three forms of degree 4. Since I is a completeintersection, we have that
i 0 1 2 3 4 5 6 7 8 9 10HF(S/I; i) 1 3 6 10 12 12 10 6 3 1 -
In particular, we have that HF(S/I; 8) 6= 0 and then #3(4, 2) = 4.
Remark 2.16. From these computations, we have that, for sum of squares inthree variables, we have three defective cases. The one coming from Alexander-Hirschowitz result in d = 1 and also d = 3 and d = 4. In terms of the de�nitiongiven in Section 2.1.2, we have that V2,2,1, which is the Veronese surface V2,2,and the varieties V2,2,3 and V2,2,4 are defective. In particular, in the last twocases we would expect that the 3rd-secant varieties would �ll the ambient space,but actually they are subvarieties of codimension 1 and 3, respectively.
In the following table, we resume the behavior of the function #2(d, 2) withrespect to d.
2.2.3 Sum of squares in four variables.
By Theorem 2.11, we know that,
#2(d, 3) ≤ 8.
By a count of parameters, we can try to guess which is the behavior of thenumerical function #2(d, 3) with respect to d. In particular, we get that
exp#2(d, 3) =
⌈dimS2d
dimSd
⌉=
⌈(2d+ 3)(2d+ 2)(2d+ 1)
(d+ 3)(d+ 2)(d+ 1)
⌉> 7⇐⇒ d > 20.
(2.2)Hence, we need to check what happen for all d ≤ 20. The idea is the same asbefore: from formula (2.2), we compute the expected 2nd-Waring rank for thegeneric form of degree 2d and then we check if it is the right answer by com-puting the Hilbert function of the corresponding generic ideal. Unfortunately,
71
# (d,2)
d1 2 3 4 5 6 ...
2
3
4
=exp # (d,2)
defe
ct
defe
ct
defe
ct
2
2
Figure 2.3: 2nd-rank of the generic forms in three variables.
since we are in four variables, the Fröberg conjecture 1 is unknown for genericideals with more than 5 generators. Consequently, we have to use the help ofCoCoA5 for the cases with expected rank bigger than 5.
From formula (2.2), we have that the expected behavior for the 2nd-Waringrank #2(d, 3) is the following.
[d = 1] This is the already known case, solved by Alexander and Hirschowitz.We are talking about quadrics in four variables and we know that it is a defec-tive case with #2(1, 3) = 4.
[d = 2] We want to know if the general quartic in four variables can bewritten as sum of 4 squares of forms of degree 2. By Theorem 2.13, we canlook at the degree 4 part of the Hilbert function of the quotient S/I whereI is a generic ideal generated by 4 general quadrics in four variables. In thisassumptions, I is a complete intersection and then we know that the Hilbertfunction is
i 0 1 2 3 4 5HF(S/I; i) 1 4 6 4 1 -
Since HF(S/I; 4) 6= 0, we have that #2(2, 3) ≥ 5. Then, we need to consider ageneric ideal I generated by 5 forms of degree 2. By Stanley's result mentionedin Section 1.1.2, we have that for such a generic ideal, the Fröberg conjecture
72
exp # (d,3)
d
2
3
4
5
6
7
8
1 2 3 4 5 6 7 8 9 10 ... 20 21 22 ...
2
Figure 2.4: Expected behavior of the function #2(d, 3) with respect to d.
holds and then we have that
HSS/I(t) =
⌈(1− t2)5
(1− t)4
⌉= 1 + 4t+ 5t2.
In particular, we have that HF(S/I; 4) = 0 and then #2(2, 3) = 5.
[d = 3] From formula 2.2, we expect that the general sextic in four variablescan be written as sum of squares of 5 cubics. Hence, we consider I to bea generic ideal generated by 5 cubics. Again, by Stanley's result mentionedabove, we have that the Fröberg conjecture holds for such I and then
HSS/I(t) =
⌈(1− t3)5
(1− t)4
⌉= 1 + 4t+ 10t2 + 15t3 + 15t4 + 6t5;
in particular, HF(S/I; 6) = 0 and then, by Theorem 2.13, #2(3, 3) = 5.
[d = 4] By formula 2.2, we expect that the general form of degree 8 can bewritten as sum of 5 squares of quartics. As above, we consider a generic idealI generated by 5 forms of degree 4 for which we can use the formula given bythe Fröberg conjecture, namely
HSS/I(t) =
⌈(1− t4)5
(1− t)4
⌉= 1 + 4t+ 10t2 + 20t3 + 30t4 + 36t5 + 34t6 + 20t7;
in particular, HF (S/I; 8) = 0 and then we have that #2(4, 3) = 5.
[d ≥ 5] In these cases, we have that the expected 2nd-Waring rank computedfrom formula (2.2) is bigger than 5. Consequently, by Theorem 2.13, we shouldcheck the vanishing of the Hilbert function in degree 2d of some quotient ring
73
S/I where I is a generic ideal with at least six generators in four variables.In those cases, Fröberg conjecture is not proven to be true, hence we need thehelp of CoCoA5 algebra software CoCoATeam [14].
As we said in Section 1.1.2, to say that S/I has Hilbert function equal to 0in degree 2d is an open property in terms of the ideal I, hence, it is enough tocheck that property on a random ideal with the same numerical character of I.
The algorithm used is the following. 1
-- Define the ambient ring.
Use S ::= ZZ/(7)[x,y,z,t];
-- Introduce the parameters:
-- D = degree the generators;
-- G = the number of generators of the ideal.
D := ;
G := ;
-- Define a random ideal generated by G forms of degree D.
F:=DensePoly(D);
L:=[];
Foreach N In 1..G Do
Randomize(F);
L:=Concat(L,[F]);
EndForeach;
I:=Ideal(L);
-- We compute the Hilbert function
Hilbert(S/I);
In conclusion, we have that, in four variables, the behavior of the 2nd-Waringrank #2(d, 3) with respect to d is given by the following table.
Remark 2.17. In this case we have two defective cases. The d = 1 case comingfrom the Alexander-Hirschowitz result which states that the Veronese threefoldV3,2 is defective in its 3rd-secant variety. Moreover, we have found that alsod = 2 is a defective case, namely the 4th-secant varietyt of V3,2,2 doesn't �ll theambient space P(S4) = P35, as we would expect.
1Since we want to check the vanishing of a Hilbert function, we can work on �nite �elds.
This helps a lot the computations especially when the degree of the generators approaches 20and the number of the generators approaches 8. For example, the computation for the ideal with7 generators in degree d = 19 over the �eld Z/7Z took around 15 minutes.
74
# (d,3)
d
2
3
4
5
6
7
8
1 2 3 4 5 6 7 8 9 10 ... 20 21 22 ...
=exp # (d,3)
defec
t
defec
t
2
2
Figure 2.5: 2nd-rank of the generic forms in four variables.
Section 2.3
A geometric approach
The method used in the previous section in the case of three and four variablesgives the answers we want, but clearly cannot be extended for any number ofvariables. To obtain an answer in the general setting we need a more theoreticalapproach.
In fact, an idea would be to investigate the geometry of the varieties Vn,k,dand their secant varieties. They are varieties of dimension
(n+dd
)− 1 and they
can be seen as images of the regular maps
νk,d,n : PSd −→ PSkd, [G] 7−→ [Gk].
Similarly to the original �Big� Waring problem, it would be enough to com-pute the dimensions of the secant varieties of these varieties Vn,k,d and thenwe simply have to see which is the �rst secant variety which �lls the ambientspace P(Skd).
Unfortunately, we have not an analogue to the Apolarity Lemma to relatethis question to the computation of the Hilbert function of some scheme in Pn;however, we may observe that such Vn,k,d is a linear projection of a Veronesevariety.
Indeed, on Sd we consider the standard monomial basisXi0,...,in := xi00 . . . xinn
for all 0 ≤ ij ≤ d such that i0 + . . . + in = d. In this way, we have that anydegree d form F ∈ Sd can be seen as a linear form in the variables X's. We uselin(F ) when we look at F as a linear form.
75
Hence, it makes sense to consider the Veronese embedding1
νN,k : P(Sd) −→ P(Symk(Sd)) where N =
(n+ d
n
);
[lin(G)] 7−→ [lin(G)k].
Moreover, we have a natural embedding of Skd inside Symk(Sd) due to basicrepresentation theory, see Landsberg [32] for more details. On the other hand,we have a linear projection of P(Symk(Sd)) over the linear subspace P(Skd).
In general, given an s-dimensional linear subspace E inside a projectivespace Pm. Let L0, . . . , Lm−s be the equations de�ning E inside Pm and let Ha (m− s)-dimensional linear subspace disjoints from E.
Then, we de�ne the linear projection from E as the map
πE : Pm −→ H ' Pm−s,[P ] 7→ [L0(P ) : . . . : Lm−s(P )].
This projection is regular on PmrE and geometrically, for any P /∈ E, πE(P )is the unique point in the intersection between the join of P with E and thelinear space H.
Now, coming back to our discussion, we have that Vn,k,d inside P(Skd) is alinear projection of VN,k from a linear subspace of P(Symk(Sd)). In order tosimplify the notations we explain it in a particular example, but the reader caneasily understand what is the generalization.
Example 2.18. Let F = a0x20 + a1x0x1 + a2x
21 be a form of degree 2 in 2
variables. Then,
ν2,2(lin(F )) =
= a20X220 + a0a1X20X11 + a0a2X20X02 + a21X
211 + a1a2X11X02 + a22X02;
but we have also,
ν1,2,2(F ) = a20x40 + a0a1x
30x1 + (a0a2 + a21)x20x
21 + a1a2x0x
21 + a22x
21.
Hence, we have that ν1,2,2(F ) = π(ν2,2(lin(F ))), where
π : P(Sym2(S2)) = P5 −→ P4 = P(S4)
[p0 : . . . : p5] 7−→ [p0 : p1 : p2 + p3 : p4 : p5].
This map is simply the projection πP of P5 from P = [0 : 0 : 1 : −1 : 0 : 0].
1In general, the polynomial ring with coe�cients in a �eld k and in n+ 1 variables, can be
seen as the symmetric algebra Sym(V ) =⊕
k∈N Symk(V ) over a k-vector space V of dimension
n + 1. This symmetric algebra is de�ned as the quotient of the tensor algebra over V modulothe symmetrizing ideal generated by the elements of the type v⊗w−w⊗ v. The gradation overthe symmetric algebra is inherited from the usual gradation of the tensor algebra.
76
In general, we have that νn,k,d(F ) = πE(νN,k(lin(F ))) where πE is the linearprojection obtained by recalling that between the degree k monomials in thevariables X's we have some relation coming from the fact that the X's areactually degree d monomials in the x's.
Hence, we have the following picture of the problem.
P(Sd)νN,k
//
νn,k,d
��
P(Symk(Sd))
πE
oo
[F ] � //_
��
[lin(F )k]
[F k]
P(Skd)
Final remarks for future work
Similarly as for the classical Waring problem considered before, the generalizedWaring problem can be reinterpreted as a question regarding the dimension ofthe secant varieties of the variety Vn,k,d. In particular, we ask
what is the minimal secant variety σs(Vn,k,d)which �lls the ambient space P(Skd)?
The answer to this question would give us the correct answer for the kth-Waringrank of generic forms of degree kd.
Hence, at this point, the idea would be to study the nature of the projectionπE described above. In particular, we want to relate the secant varieties of theVeronese variety VN,k with the secant varieties of the variety Vn,k,d.
Proposition 2.19. With the same notation as above,
1. πE is regular over VN,k;
2. Vn,k,d = πE(VN,k);
3. σs(Vn,k,d) ⊃ πE(σs(VN,k)).
Proof. (1) We know that a linear projection πE is regular outside the linearspace E; hence, it is enough to show that VN,k is disjoint from E. We knowthat the image of a point [lin(F )k] ∈ VN,k along the map πE is the point [F k].Now, if [lin(F )k] was on E, we have that F k ≡ 0; thus, it has to be F ≡ 0.Consequently, lin(F ) ≡ 0.
(2) By de�nition, we have that πE(VN,k) ⊂ Vn,k,d. Now, given a point[F k] ∈ Vn,k,d, we can simply replace the degree d monomials of F with thecorresponding X-variables; hence, we get a point [lin(F )k] ∈ Vn,k,d which ismapped to [F k] via πE .
(3) Let [P ] = [lin(F1)k] + . . .+ [lin(Fs)k] be a generic point in σs(VN,k). By
de�nition, we have that [πE(P )] = [F k1 ] + . . .+ [F ks ] ∈ σs(Vn,k,d).
77
From these observations, we can directly observe that the dimension of thesth-secant variety of Vn,k,d is bounded from below by the dimension of the sth-secant variety of the Veronese variety VN,k. By Alexander-Hirschowitz theorem,in particular by Corollary 2.9, we know the dimension of secant varieties of theVeronese variety VN,k; hence, the �rst question we should ask is
what is the �ber of the projection πE?
With the answer of that question, we could understand when the projection ofthe sth-secant variety of VN,k �lls the whole P(Skd). By Proposition 2.19(3),that number s would be an upper bound for our kth-Waring Waring rank.
Afterwards, to produce the correct answer to our problem, we could try toanswer to the question
is σs(Vn,k,d) = πE(σs(VN,d))?
In case of positive answer, by knowing the dimension of the projection of se-cant varieties of Veronese variety, we would have the complete answer of ourgeneralized Waring problem.
Section 2.4
Monomials as sum of kth-powers (joint work with E. Carlini 1)
Let S :=⊕
i∈N Si = C[x0, . . . , xn] be the ring of polynomials in n+ 1 variableswith complex coe�cients and with the standard gradation. The aim of thissection is to consider the generalized Waring problem introduced in Fröberg,Ottaviani, and Shapiro [25] and described in Section 2.2 in the case of mono-
mials.
2.4.1 Basic facts
In Carlini, Catalisano, and Geramita [11], the authors gave an explicit formulato compute the Waring rank of a given monomial in any number of variablesand any degree.
Theorem 2.20 (Carlini, Catalisano, and Geramita [11]). Given a monomial
M = xa00 . . . xann of degree k such that 1 ≤ a0 ≤ . . . ≤ an, then
#k(M) =1
a0 + 1
n∏i=0
(ai + 1). (2.3)
The idea is to see how this result can help us to compute the kth-Waring rankof a monomial M of degree kd.
First, we introduce some elementary tools to study.
1School of Mathematical Sciences, Monash University. Melbourne, AUSTRALIA
E-mail address: [email protected] accepted in Communications in Algebra.
78
Remark 2.21. Consider a monomialM of degree kd in the variables {x0, . . . , xn}.We say that a monomial M ′ of degree kd′ in the variables {X0, . . . , Xm} is agrouping ofM if there exists a positive integer l such that d = ld′ andM can beobtained from M ′ by substituting each variable Xi with a monomial of degreel in the x's, i.e. Xi = Ni(x0, . . . , xn) for each i = 1, . . . ,m with deg(Ni) = l.The relation between the kth-rank of M and M ′ is given by
#k(M ′) ≥ #k(M).
Indeed, given a decomposition of M ′ as sum of kth-powers, i.e.
M ′ =
r∑i=1
Fi(X0, . . . , Xn)k, with deg(Fi) = d′,
we can write a decomposition for M by using the substitution given above, i.e.
M =
r∑i=1
Fi(N0(x0, . . . , xn), . . . , Nm(x0, . . . , xn))k.
Remark 2.22. Consider a monomialM of degree kd in the variables {x0, . . . , xn}.We say that a monomial M ′ of the same degree is a specialization of M if M ′
can be found from M after a certain number of identi�cations of the typexi = xj . Again, it makes sense to compare the two kth-ranks and we get
#k(M) ≥ #k(M ′).
Indeed, given a decomposition of M as sum of r kth-powers, we can writea decomposition for M ′ with the same number of summands applying theidenti�cations between variables to each addend.
Remark 2.23. Consider a monomial M of degree kd1 and N a monomial ofdegree d2. We can look at the monomial M ′ = MNk. Clearly the degree of M ′
is also divisible by k; again, it makes sense to compare the kth-rank of M andM ′. The relation is
#k(M) ≥ #k(M ′).
Indeed, given a decomposition as sum of kth-powers for M , e.g. M =∑ri=1 F
ki
with Fi's forms of degree d1, we can easily �nd a decomposition for M ′ withthe same number of summands, i.e. M ′ = MNk =
∑ri=1(FiN)k.
The inequality on the kth-rank in Remark 2.23 can be strict in general as wecan see in the following example.
Example 2.24. Consider k = 3 and the monomials M = x1x2x3 and M ′ =(x20)3M = x60x1x2x3. By Theorem 2.20, we know that #3(M) = #3(x1x2x3) =4, but we can consider a grouping of the monomialM ′, i.e.M ′ = (x30)2(x1x2x3) =X2
0X1. By Remark 2.21 and Theorem 2.20, we have #3(M ′) ≤ #3(X20X1) = 3.
As a straightforward application of these remarks we get the following lemmawhich is useful to reduce the number of cases to consider once k and n are�xed.
79
Lemma 2.25. Given a monomial M = xa00 xa11 · · ·xann of degree kd, then
#k(M) ≤ #k([M ]),
where [M ] := x[a0]k0 x
[a1]k1 · · ·x[an]kn , where the [ai]k's are the remainders of the
ai's modulo k.
Proof. We can write ai = kαi + [ai]k for each i = 0, . . . , n. Hence, we getthat M = Nk[M ], where N = xα0
0 xα11 · · ·xαn
n . Obviously, k|deg([M ]) and byRemark 2.23, we are done.
Remark 2.26. With the above notations and numerical assumptions, we havethat [a0]k+ · · ·+[an]k is a multiple of k and also it has to be at most (k−1)(n+1) = kn−n+ k− 1. Hence, �xed the number of variables n+ 1 and the integerk, we will have to consider only a few cases with respect to the remainders ofthe exponents modulo k.
2.4.2 Results on the kth-rank for monomials
In this section we collect our results on the k-th rank of monomials.
General case
We begin by presenting some general results on the k-th Waring rank for mono-mials.
Remark 2.27. Using the idea of grouping variables, we can easily get a com-plete description of the k = 2 case. Given a monomial M of degree 2d which isnot a square, we have #2(M) = 2. Indeed,
M = XY =
[1
2(X + Y )
]2+
[i
2(X − Y )
]2,
where X and Y are two monomials of degree d.
In the next result, we see that case k = 2 is the unique case in which thekth-rank of a monomial can be equal to two.
Theorem 2.28. If M is a monomial of degree kd, then #k(M) ≤ 2k−1. More-
over, #k(M) = 2 if and only if k = 2 and M is not a square.
Proof. Any monomialM ∈ Skd is a specialization of the monomial x1 · . . . ·xkd.Now, we can consider the grouping given by
X1 = x1 · . . . · xd, . . . , Xk = x(k−1)d+1 · · · . . . · xkd.
Thus, by Remark 2.22, Remark 2.21 and Theorem 2.20, we get the bound
#k(M) ≤ #k(X1 · . . . ·Xk) = 2k−1.
80
Now suppose that k > 2 and #k(M) = 2. Hence, we can writeM = Ak−Bkfor suitable A,B ∈ Sd. Factoring we get
M =
k∏i=1
(A− ξiB),
where the ξi are the kth-roots of 1. In particular, the forms A− ξiB are mono-
mials. If M is not a kth-power, using A − ξ1B and A − ξ2B we get that Aand B are not trivial binomials. Hence a contradiction as A − ξ3B cannot bea monomial. To conclude the proof we use the k = 2 case seen in Remark2.27.
Remark 2.29. For n ≥ 2 and k small enough, we may observe that our resultgives a better upper bound for the kth-rank of monomials of degree kd thanthe general result of Fröberg, Ottaviani, and Shapiro [25]. Indeed, if we lookfor which k the inequality 2k−1 ≤ kn holds, for n = 2 we have k ≤ 6 and, forn = 3, k ≤ 9. Increasing n, we can �nd even better results, e.g. for n = 10 ourTheorem 2.28 gives a better upperbound (for monomials) for any k ≤ 59.
Two variables case (n = 1).
In the case of binary monomials, we can improve the upper bound given inTheorem 2.28.Proposition 2.30. Let M = xa00 x
a11 be a binary monomial of degree kd. Then,
#k(M) ≤ max{[a0]k, [a1]k}+ 1.
Proof. By Lemma 2.25, we know that #k(M) ≤ #k([M ]); hence, we consider
the monomial [M ] = x[a0]k0 x
[a1]k1 . Now, we observe that, as we said in Remark
2.26, the degree of [M ] is a multiple of k and also ≤ 2k− 2; hence, deg([M ]) iseither equal to 0, i.e. [M ] = 1, or k. In the �rst case, it means that M was apure kth-power, and the kth-rank is
#k(M) = 1 = max{[a0]k, [a1]k}+ 1.
If deg([M ]) = k, we can apply Theorem 2.20 to [M ] and we get
#k(M) ≤ #k([M ]) = max{[a0]k, [a1]k}+ 1.
Remark 2.31. As a consequence of Proposition 2.30, for binary monomialswe have that #k(M) ≤ k. Actually, this upper bound can be directly derivedfrom the main result in Fröberg, Ottaviani, and Shapiro [25]. We observe thatthis upperbound is sharp by considering #k(x0x
k−11 ) = k.
As a consequence of Theorem 2.28, we are able to easily give a solution for thek = 3 case for binary monomials.
Corollary 2.32. Given a binary monomial M of degree 3d, we have
81
1. #3(M) = 1 if M is a pure cube;
2. #3(M) = 3 otherwise.
Proof. By Remark 2.31, we have that the 3rd-rank can be at most 3; on theother hand, by Theorem 2.28, we have that, M is not a pure cube, the rankhas to be at least 3.
For k ≥ 4 the situation is not so easily described and even in the case k = 4we have only partial results.
Remark 2.33. The �rst new step is to consider the k = 4 case for binarymonomials. In such case we can only have rank 1, 3 or 4.
Let M = xa00 xa11 be a binary monomial of degree 4d. By Remark 2.25, we
can consider the monomial [M ] obtained by considering the exponents modulo4. Since [M ] has degree divisible by 4 and less or equal to 6, we have to consideronly three cases with respect the remainders of the exponents modulo 4, i.e.
([a0]4, [a1]4) ∈ {(0, 0), (1, 3), (2, 2)}.
The (0, 0) case corresponds to pure fourth powers, i.e. monomials with 4th-rank equal to 1. In the (2, 2) case we have
#4(M) ≤ #4(x20x21) = 3;
since the 4th-rank cannot be two, we have that binary monomials in the (2, 2)class have 4th-rank equal to three.
Unfortunately, we cannot conclude in the same way the (1, 3) case. Since#4(x0x
31) = 4, a monomial in the (1, 3) class could still have rank equal to 4.
Indeed, for example, by using the computer algebra system CoCoA, we havecomputed #4(x0x
71) = #4(x30x
51) = 4.
A similar analysis can be performed for k ≥ 5, but we can only obtain partialresults.
k = 3 case in three and more variables.
In this section we consider the case k = 3 with more than two variables. ByTheorem 2.28, we have that, also in this case, we can only have 3rd-rank equalto 1, 3 or 4.
This lack of space allows us to give a complete solution for monomials inthree variables and degree 3d.
Proposition 2.34. Given a monomial M = xa00 xa11 x
a22 of degree 3d, we have
that
1. #3(M) = 1 if M is a pure cube;
2. #3(M) = 4 if M = x0x1x2;
3. #3(M) = 3 otherwise.
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Proof. By Lemma 2.25, we consider the monomials [M ] with degree divisibleby 3 and less or equal than 6. Hence, we have only four possible cases, i.e.
([a0]3, [a1]3, [a2]3) ∈ {(0, 0, 0), (0, 1, 2), (1, 1, 1), (2, 2, 2)}.
The (0, 0, 0) case corresponds to pure cubes and then to monomials with 3rd-rank equal to one. In the (0, 1, 2) case we have, by Theorem 2.20,
#3(M) ≤ #3(x1x22) = 3;
since, by Theorem 2.28, the rank of monomials which are not pure cubes is atleast 3, we get the equality. Similarly, we conclude that we have rank three alsofor monomials in the (2, 2, 2) class. Indeed, by using grouping and Theorem2.20, we have
#3(M) ≤ #3(x20x21x
22) ≤ #3(XY 2) = 3.
Now, we just need to consider the (1, 1, 1) class.By Theorem 2.20, we have #3(x0x1x2) = 4. Hence, we can consider mono-
mials M = xa00 xa11 x
a22 with a0 = 3α + 1, a1 = 3β + 1, a2 = 3γ + 1 and where
at least one of α, β, γ is at least one, say α > 0. By Remark 2.23, we have
#3(M) = #3((xα−10 xβ1xγ2)3x40x1x2) ≤ #3(x40x1x2).
Now, to conclude the proof, it is enough to show that #3(x40x1x2) = 3. Indeed,we can write
x40x1x2 =
[√1
6x20 + x1x2
]3+
[−1
6x20 + x1x2
]3+[
3√−2x1x2
]3,
and thus we are done.
Using the same ideas, we can produce partial results in the four and �ve vari-ables cases with k = 3.
Remark 2.35. Given a monomial M = xa00 xa11 x
a22 x
a33 with degree 3d, we
consider the monomial [M ] which has degree divisible by 4 and less or equalthan 8. Hence, we need to consider only the following classes with respect tothe remainders of the exponents modulo 3
([a0]3, [a1]3, [a2]3, [a3]3) ∈ {(0, 0, 0, 0), (0, 0, 1, 2), (0, 1, 1, 1), (0, 2, 2, 2), (1, 1, 2, 2)}.
The (0, 0, 0, 0) case corresponds to pure cubes and we have rank equal to one.Now, we use again Lemma 2.25, grouping and Theorem 2.20.In the (0, 0, 1, 2) class, we have
#3(M) ≤ #3(x2x23) = 3;
in the (0, 2, 2, 2) class, we have
#3(M) ≤ #3(x21x22x
23) ≤ #3(XY 2) = 3;
83
in the (1, 1, 2, 2) class, we have
#3(M) ≤ #3(x0x1x22x
23) ≤ #3((x0x1)(x2x3)2) = #3(XY 2) = 3.
Again, since the 3rd-rank has to be at least three by Theorem 2.28, we concludethat in these classes the 3rd-rank is equal to three.
The (0, 1, 1, 1) class is a unique missing case because the upper bound with#3(x1x2x3) = 4 is clearly useless. Another idea would be to compute the 3rd-rank of x30x1x2x3. Indeed, each monomial in four variables and degree 3d is ofthe type Nk(x30x1x2x3), hence, by Remark 2.23, we have
#3(M) ≤ #3(x30x1x2x3).
Finding #3(x30x1x2x3) = 3, we would be done.
Remark 2.36. Given a monomial M = xa00 xa11 x
a22 x
a33 x
a44 with degree 3d, we
consider the monomial [M ] which has degree divisible by 4 and less or equal to10. Hence, we need to consider only the following classes with respect to theremainders of the exponents modulo 3.
The (0, 0, 0, 0, 0) class corresponds to pure cubes and 3rd-rank equal to one.By using Lemma 2.25, grouping, previous results in three or four variables andTheorem 2.20, we get the following results.
In the (0, 0, 0, 1, 2) case, we have
#3(M) ≤ #3(x3x24) = 3;
in the (0, 0, 2, 2, 2) case, we have
#3(M) ≤ #3(x21x22x
23) = 3;
in the (0, 1, 1, 2, 2) case, we have
#3(M) ≤ #3(x0x1x22x
23) = 3;
in the (1, 2, 2, 2, 2) case, we have
#3(M) ≤ #3(x0x21x
22x
23x
24) = #3((x0x
21)(x2x3x4)2) ≤ #3(XY 2) = 3.
Hence, by Theorem 2.28, in these cases we have 3rd-rank equal to three.
There are only two missing cases: the (0, 0, 1, 1, 1) case, which can be reducedto the unique missing case in four variables seen above; the (1, 1, 1, 1, 2) case,for which it would be enough to show that #3(x0x1x2x3x
24) = 3.
2.4.3 Final remarks
We conclude with some �nal remarks which suggest some projects for the fu-ture.
84
Remark 2.37. In this paper we work over the �eld of complex numbers.However, for a monomial M ∈ Skd it is reasonable to look for a real Waringdecomposition, i.e. M =
∑F ki where each Fi has real coe�cients. Even if
Remarks 2.21,2.22, and 2.23 still hold over the reals, this is not longer true forTheorem 2.20. This is the main obstacle to extend our results over R. However,in Boij, Carlini, and Geramita [9] it is shown that the degree d monomial xayb
is the sum of a+b, and no fewer, d-th powers of real linear forms. Thus, we caneasily prove the analogue of Proposition 2.30. Let #k(M,R) be the real k-thrank, then
#k(x0a0x1
a1 ,R) ≤ [a0]d + [a1]d
and the bound is sharp. Notice that Corollary 2.32 cannot be extended to thereal case as we cannot use Theorem 2.28.
Remark 2.38. In Carlini, Catalisano, and Geramita [11] it is proved thatmonomials in three variables produce example of forms having (standard) War-ing rank higher than the generic form. This is no longer true, in general, forthe k-th rank. For example, in the k = 3 case and in three variables, the 3-rdrank of a monomial is at most 4. While, for d� 0 the 3-rd rank of the genericform of degree 3d is 9, see Fröberg, Ottaviani, and Shapiro [25].
85
86
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