Università degli Studi di Pisa
Facoltà di Scienze Matematiche, Fisiche e Naturali
Corso di Laurea Magistrale in Matematica
On the Hilbert quasi-polynomials of
non-standard graded rings
Candidato:
Carla Mascia
Relatore:
Dott. Massimo Caboara
Controrelatore:
Prof.ssa Patrizia Gianni
Contents
Introduction v
1 Basic knowledge 1
1.1 Graded rings and graded modules . . . 1
1.2 Polynomial rings graded by matrices . . . 3
1.3 Hilbert function and Hilbert-Poincaré series . . . 5
1.4 Hilbert-Serre theorem and Hilbert polynomial . . . 9
2 Hilbert quasi-polynomial 15 2.1 Existence of the Hilbert quasi-polynomial . . . 15
2.2 Degree and leading coefficient of the Hilbert quasi-polynomial 20 2.3 Proprieties of the coefficients of the Hilbert quasi-polynomial 28 2.4 Formulas for the other coefficients of the Hilbert quasi-polynomial . . . 32
3 Algorithm for computing Hilbert quasi-polynomials 45 3.1 Algorithm implementation . . . 45
3.2 Numerical experiments . . . 50
4 Multigraded Hilbert-Poincaré series 55 4.1 Simplicial complexes and corners . . . 55
4.2 The Slice Algorithm . . . 56
4.3 Corner-Euler Algorithm . . . 59
4.4 Computing Hilbert-Poincaré series . . . 61
Conclusions 65
A System of linear Diophantine equations and Smith normal
form 67
B Code of the algorithm for computing Hilbert
quasi-polynomials 71
Bibliography 79
Introduction
The Hilbert function, its generating function and the Hilbert polynomial of a graded R-module M have been extensively studied since the famous pa-per of Hilbert: Ueber die Theorie der algebraischen Formen ([Hilbert, 1890]). In particular the coefficients and the degree of the Hilbert polynomial play an important role in Algebraic Geometry, e.g. they are an efficient way for computing the dimension and the degree of an algebraic variety defined by explicit polynomial equations.
The main aim of this thesis is to generalize the well-know theory of the Hilbert polynomial for standard graded rings to the general case of rings graded by any vector in Nk
+. The Hilbert function of a non-standard graded
ring is of quasi-polynomial type. We investigate the structure of Hilbert quasi-polynomials, such as degree, leading coefficient and proprieties of their coefficients, and at the same time we present an algorithm to calculate it.
The Hilbert quasi-polynomials have application in several fields, for ex-ample Bruns and Ichim in [Bruns, 2007] have used Hilbert quasi-polynomials to give a purely algebraic proof of an old combinatorial result due to Ehrhart, McMullen and Stanley.
The computation of Hilbert-Poincaré series has also received a lot of at-tention. We consider the problem to get only a part of the Hilbert-Poincaré series, without computing it entirely, and we propose a solution by means of an algorithm due to Roune ([Roune, 2010]) which exploit the concepts of corners and Koszul simplicial complexes, originally introduced by Bayer ([Bayer, 1996]).
Let us see in more detail the main results of this thesis.
Let R := K[x1, . . . , xk]be a polynomial ring graded by a positive matrix
W ∈Matm,k(N) and let I be an ideal of R homogeneous respect to W . The
numerical function HR/I : Nm → N defined by HR/I(d) := dimK((R/I)d) is
called Hilbert function of R/I.
If the graduation on R is standard (i.e. W = [1, . . . , 1] ∈ Nk), H R/I is
definitively equal to a polynomial, called Hilbert polynomial of R/I, which has degree less or equal to k − 1.
We focus our attention on the less well-known case of non-standard grad-v
vi CONTENTS uation.
Let R be graded by W := [d1, . . . , dk] ∈ Nk+. The Hilbert function is
definitely equal to a quasi-polynomial PW
R/I := {P0, . . . , Pd−1} of period d,
where d := lcm(d1, . . . , dk), that is HR/I(n) = Pi(n) for all n sufficiently
large and n ≡ i (mod d).
We prove the existence of Hilbert quasi-polynomials and we give an ex-pository account of their construction. For this purpose, a particular power series associated to the Hilbert polynomial, called Hilbert-Poincaré series, plays a key role. It is defined by HPR/I(t) :=
P
n≥0HR/I(n)tn.
We show that PW
R/I is a quasi-polynomial with rational coefficients and,
for all i = 0, . . . , d − 1, if I = (0) then Pi has degree equal to k − 1, else Pi
has degree less or equal to k − 2. Anyway, for each Pi
lc (Pi) =
1 (k − 1)!Qk
i=1di
We restrict our study to (R/I, W ) with I = (0) and W = [d1, . . . , dk]
a weight vector such that gcd(d1, . . . , dk) = 1, in fact we can easily obtain
PR/IW0 by PRW, where W = gcd(d01 1,...,d0k)
W0.
We present some proprieties of the coefficients of PW
R . Through an
anal-ysis of some numerical examples, we have noted some regularity for the coef-ficients of PW
R . In particular, we prove that PRW can be written as a sum of a
polynomial with rational coefficients of degree k − 1 and a quasi-polynomial with rational coefficients of degree δ − 1, where
δ := max{|I| | gcd(di)i∈I 6= 1and I ⊆ {1, . . . , k}}
This means that the coefficients of terms of degree δ, . . . , k − 1 are the same for all Pi. Whereas, the r-th coefficient of PRW has period equal to
δr := lcm( gcd(di)i∈I | |I| = r + 1and I ⊆ {1, . . . , k})
We devise a strategy to find the formulas for the coefficients of Hilbert quasi-polynomials and we compute two of them. If we denote the rth coef-ficient of PW
R by cr and if we suppose δ ≤ r , we have
ck−2 = Pk i=1di 2(k − 2)!Qk i=1di ck−3= 3(Pk i=1di)2−Pki=1d2i 24(k − 3)!Qk i=1di
In addition, we present an algorithm for the effective calculation of Hilbert quasi-polynomials. We have implemented it in SINGULAR, a free and
open-source computer algebra system for polynomial computations, and we show some numerical experiments.
CONTENTS vii In the last part of this thesis we consider an other related problem: for some graduations on R := K[x1, . . . , xk]given by W = (wij) ∈Matm,k(N),
the computation of Hilbert-Poincarè series associated to R/I is very expen-sive. For this reason, we present a way to get only a part of Hilbert-Poincaré series without computing it entirely. We use the following formula for the Hilbert-Poincaré series which exploits corners and Koszul simplicial com-plexes of I
H(I) := 1 + X
m∈cor(I)
χ(∆Im) m
Here, χ denotes the Euler characteristic of a simplicial complex, ∆I m is the
Koszul simplicial complex of I at monomial m, and cor(I) is the set of corners of I. By substituting tw1i
1 · · · twmmi for xi in H(I), for all i = 1, . . . , k,
we obtain exactly the numerator f(t1, . . . , tm)of the Hilbert-Poincaré series.
If we want only a part of f(t1, . . . , tm), we stop the computation of corners
at one point, instead of computing all corners, in order to reduce the cost of computation.
This thesis expands Cirillo’s bachelor thesis ([Cirillo, 2014]), in which he shows only the existence of Hilbert quasi-polynomials and he gives a simple algorithm implemented in CoCoA to compute them.
This thesis is structured as follows:
Chapter 1: An overview of Hilbert function, Hilbert-Poincaré series and Hilbert polynomial of standard graded rings.
Chapter 2: Our results concerning Hilbert quasi-polynomial, such as exis-tence, degree, proprieties and formula for its coefficients.
Chapter 3: An algorithm for computing Hilbert quasi-polynomial.
Chapter 4: A strategy to get a part of Hilbert-Poincaré series without com-puting it entirely.
Chapter 1
Basic knowledge
In this chapter we recall basic definitions, as graded rings and graded modules, and some well known results concerning Hilbert function, Hilbert-Poincaré series and Hilbert polynomial associated to a graded module. Most of the material in this chapter is standard and we refer to [Robbiano, 2005] for the proofs and for further results.
In this work, all monoids and rings are assumed to be commutative.
1.1
Graded rings and graded modules
In this section we introduce and study quite general kinds of gradings, namely graded rings and graded modules.
Definition 1.1Let (Γ, +) be a monoid. A ring R is called a graded ring over Γ, or a Γ-graded ring, if there exists a family of additive subgroups {Ri}i∈Γ such that
• R = Li∈ΓRi,
• RiRj ⊆ Ri+j for all i, j ∈ Γ.
The elements of Ri are called homogeneous elements of degree i.
The subgroup Ri is called the homogeneous component of degree i of
R.
We note that, if R is a Γ-graded ring and 0 is the identity of R, then 0 is a homogeneous element of R of every degree. Moreover, the decomposition of every element into its homogeneous components is unique, since in the definition there is a direct sum.
2 CHAPTER 1. BASIC KNOWLEDGE The following two examples constitute the most important situations in which we shall meet graded rings.
Example 1.2. Let S be a ring, let n ≥ 1, and let R = S[x1, . . . , xn] be a
polynomial ring over S. If we let
Ri := {f ∈ R | deg(t) = i for all t ∈ Supp(f)}
for every i ∈ N, we make R into an N-graded ring, where deg(xα1
1 · · · x αn
n ) := α1+ · · · + αn
This grading satisfies deg(x1) = · · · = deg(xn) = 1and is called the
stan-dard gradingof R. The elements of Ri are called homogeneous
polyno-mialsof degree i.
Example 1.3. Let S be a ring, let n ≥ 1, and let R = S[x1, . . . , xn] be
a polynomial ring over S. For each α := (α1, . . . , αn) ∈ Nn, we define
Rα := S · xα11· · · xnαn. In this way R = Lα∈NnRα become an Nn-graded
ring. We can also view R as a Zn-graded ring if we define R
α = 0 for all
α = (α1, . . . , αn) ∈ Zn such that αi< 0 for some i ∈ {1, . . . n}.
Throughout the remainder of this chapter, R will be a Γ-graded ring, where (Γ, +) denotes a monoid.
Definition 1.4A R-module M is called a Γ-graded module over R, or a Γ-graded R-module, if there exists a family of subgroups {Md}d∈Γ such
that
• M = Ld∈ΓMd,
• RiMd⊆ Mi+d for all i, d ∈ Γ.
The elements of Mdare called homogeneous elements of degree d of M.
The submodule Md is called the homogeneous component of degree d
of M.
Definition 1.5An R-submodule N of the Γ-graded R-module M is called a Γ-graded R-submodule of M if we have N = Ld∈Γ(N ∩ Md).
A Γ-graded submodule of R is also called a Γ-homogeneous ideal of R. Let N ⊆ M be a Γ-graded R-submodule. We can equip the residue class module M/N with the structure of a Γ-graded R-module by defining
1.2. POLYNOMIAL RINGS GRADED BY MATRICES 3 (M/N )d:= Md/Ndfor every d ∈ Γ. In particular, the residue class ring R/I
of R by a homogeneous ideal I is again a Γ-graded ring.
Given a Γ-graded R-module M, there exists a notation to indicate that we have altered M by shifting its grading by γ steps.
Definition 1.6Let M be a Γ-graded module over a Γ-graded ring R with graduation given by M := {Md}d∈Γ and let γ ∈ Γ. Then, the family
{M0
d}d∈Γ := Md+γ is still a graduation of M and it is the graduation M
shifted by γ. The shift of the graduation M by γ is denoted by M(γ). Now we present the sets of morphisms between Γ-graded R-modules. Definition 1.7
(i) Let M, M0 be two Γ-graded R-modules. A morphism f : M → M0 is
called a morphism of degree γ ∈ Γ if f(Md) ⊆ Md+γ0 for all d ∈ Γ.
In particular, if γ = 0, f is called a homogeneous morphism. (ii) An exact sequence of graded modules is called graded of degree γ if
all its morphism are graded of degree γ. In particular, if γ = 0, it is called a homogeneous exact sequence.
1.2
Polynomial rings graded by matrices
Having introduced the general notions of graded rings and modules, we are now going to concentrate on certain specific kinds of graded object. We are interested in gradings on polynomial rings which have two properties: the indeterminates are homogeneous and the constants are homogeneous of degree zero.
In what follows, let K be a field, n ≥ 1, and R := K[x1, . . . , xk]be a
polyno-mial ring over K.
Definition 1.8Let m ≥ 1 and let R be equipped with a Zm-grading such that K ⊆ R0 and the indeterminates x1, . . . , xk are homogeneous elements.
For j = 1, . . . , k, let (w1j, . . . , wmj) ∈ Zm be the degree of xj. The matrix
W = (wij) ∈ Matm,k(Z) is called the degree matrix of the grading. In
other words, the columns of the degree matrix are the degrees of the inde-terminates. The rows of the degree matrix are called the weight vectors of the indeterminates x1, . . . , xk.
4 CHAPTER 1. BASIC KNOWLEDGE Let R be graded by a matrix W ∈ Matm,k(Z), then the degree of a term
t = xα1
1 · · · x αk
k of R is given by degW(t) := W · (α1, . . . , αk)T. The vector
(α1, . . . , αk)T is usually denoted by log(t).
In addition, the set of homogeneous polynomials of degree d ∈ Zmis denoted
by RW
d , or simply by Rd if it is clear which grading we are considering. A
polynomial f ∈ RW
d is also called homogeneous of degree d.
Example 1.9. Let R = K[x, y, z] be graded by the matrix W =1 2 4
0 1 0
and consider the polynomial f = x4y + yz. Then f is homogeneous of degree
(6, 1), because W · log(x4y)T = W · log(yz)T = (6, 1).
In view of the following proposition, we shall henceforward assume tacitly that the degree matrix W has full rank.
Proposition 1.10Let R be graded by a matrix W ∈ Matm,k(Z), W 6= 0, and let W have Z-linearly dependent rows, and let W0 be a submatrix of W
which consists of a maximal linearly independent set of rows. Denote the number of rows of W0 by m0, and let V ∈ Mat
m,m0(Z) be the matrix such
that V · W0= W. Then it holds RW0
d0 = R W
V ·d for all d0∈ Zm
0
.
Our next goal is to find hypothesis which force the homogeneous com-ponents of R to be finite dimensional K-space vectors, in order to define the Hilbert function of R which describes exactly the dimension of homogeneous components of R as K-space vectors.
Definition 1.11Let m ≥ 1, let R be graded by a matrix W of rank m in Matm,k(Z), and let w1, . . . , wm be the rows of W . We say that the grading
on R given by W is of positive type if there exist a1, . . . , am ∈ Z such that
all entries of a1w1+ · · · + amwm are positive. In this case, we shall also say
that W is a matrix of positive type.
Example 1.12. The standard grading of R, defined in Example 1.2, is given by (1, . . . , 1) ∈ Zk and it is of positive type.
Proposition 1.13Let R be graded by a matrix W ∈ Matm,k(Z) of positive
type, and let M be a finitely generated graded R-module. Then (i) R0 = K
1.3. HILBERT FUNCTION AND HILBERT-POINCARÉ SERIES 5 (ii) dimK(Md) < ∞, for all d ∈ Zm.
Definition 1.14Let W ∈ Matm,k(Z) be a matrix of rank m. The grading
on R defined by W is called positive if no column of W is zero and the first non-zero element in each column is positive. In this case, we shall also say that W is a positive matrix.
Notice that if W is positive, then we have degW(xi) > 0 for i = 1, . . . , k,
and hence R+=Ld>0RdW = (x1, . . . , xk).
The following proposition says that positive gradings are of positive type and there exists a change of grading which transforms a grading of positive type into a positive one.
Proposition 1.15 Let R be graded by a matrix W ∈ Matm,k(Z) (i) If the grading defined by W is positive, it is of positive type.
(ii) If the grading defined by W is of positive type, then there exists a non-singular matrix V ∈ Matm(Z) such that the grading defined by
the matrix W0 = V · W is positive and RW0
V ·d = RWd for all d ∈ Zm.
From now on, we suppose R graded by a positive matrix W ∈ Matm,k(Z).
1.3
Hilbert function and Hilbert-Poincaré series
We are going to present two strongly related notions: the Hilbert function and the Hilbert-Poincaré series of a finitely generated graded module, which measure the growth of the dimension of its homogeneous components. Definition 1.16Let M be a finitely generated graded R-module. The nu-merical function HW
M : Zm−→ N defined by
HMW(d) := dimK(Md)
is called the Hilbert function of M.
When the graduation given by W is clear from the contest, we denote simply Hilbert function by HM instead of HMW.
6 CHAPTER 1. BASIC KNOWLEDGE Proposition 1.17 (Basic properties of Hilbert functions)
Let M be a finitely graded R-module.
(i) For every d ∈ Zm, the Hilbert function of the shifted module M(d) is
given by
HM (d)(i) = HM(d + i)
for all i ∈ Zm
(ii) Given finitely generated graded R-modules M0, M00and a homogeneous
short exact sequence of graded R-modules 0 → M0 → M → M00 → 0,
it holds HM(i) = HM0(i) + HM00(i)for all i ∈ Zm.
(iii) Given finitely many finitely generated R-modules M1, . . . , Mr, it holds
HM1⊕···⊕Mr(i) = HM1(i) + · · · + HMr(i)
for all i ∈ Zm.
Remark 1.18. Let W = (wij) ∈Matm,k(Z) and (i1, . . . , im) ∈ Zm. Then the
value HR(i1, . . . , im)of the Hilbert function of R graded by W is the number
of solutions (α1, . . . , αk) ∈ Nk of the system of Diophantine equations
w11y1+ · · · + w1kyk = i1 w21y1+ · · · + w2kyk = i2 ... ... wm1y1+ · · · + wmkyk = im (1.1) in the indeterminates y1, . . . , yk.
In appendix A, we recall how to solve a system of linear Diophantine equations by means of Smith normal form. This method produces all in-teger solutions, from which we have to consider only the positive ones. In general, finding the solutions of the system (1.1) is not the best way to com-pute the value of HR(i1, . . . , im) for a given (i1, . . . , im) ∈ Zm. Admittedly,
the description of HR through the solutions of system (1.1) is not very
il-luminating with respect to the overall structure of this function. To get a deeper insight, we use the values of the Hilbert function as the coefficients of a suitable extended power series.
Actually, in the standard graded case, there is a simple formula for HR
(Proposition (1.21)). Whereas, for m = 1 the problem is equivalent to determine the formula for the number of partitions of an integer into elements of a finite set and we’ll examine this case closer in Chapter 3.
1.3. HILBERT FUNCTION AND HILBERT-POINCARÉ SERIES 7 Definition 1.19Let M be a finitely generated graded R-module. The ex-tended power series
HPMW(t) := X
d∈Nm
HMW(d)td∈ NJtK
where t = (t1, . . . , tm), is called the Hilbert-Poincaré series of M.
In combinatorial terms, the Hilbert-Poincaré series of M is the generating function of the sequence given by the Hilbert function.
When the graduation given by W is clear from the contest, we denote simply Hilbert-Poincaré series by HPM instead of HPMW.
Proposition 1.20 (Basic properties of Hilbert-Poincaré series) Let M be a finitely generated graded R-module.
(i) For every d = (d1, . . . , dm) ∈ Zm, the Hilbert-Poincaré series of the
shifted module M(d) is given by
HPM (d)(t1, . . . , tm) = t−d1 1· · · t−dmmHPM(t1, . . . , tm)
(ii) Given finitely generated graded R-modules M0, M00and a homogeneous
short exact sequence of graded R-modules 0 → M0 → M → M00 → 0,
it holds HPM(t) = HPM0(t) + HPM00(t).
(iii) Given finitely many finitely generated R-modules M1, . . . , Mr, it holds
HPM1⊕···⊕Mr(t) = HPM1(t) + · · · + HPMr(t)
(iv) Given a homogeneous polynomial f ∈ R\{0}, it holds
HPM/f M(t) = HPM(t) − td11· · · tmdmHPM/(0:M(f ))(t) (1.2)
where d = (d1, . . . , dm) ∈ Zm is the degree of f.
In particular, if f is a non-zerodivisor for M then M/(0 :M (f )) = M
and hence
HPM/f M(t) = (1 − td11· · · t dm
m )HPM(t) (1.3)
Equation (1.2) is know as main formula and it is a very useful tool to compute Hilbert-Poincaré series of M. We observe that, if I is a homoge-neous ideal of R and f ∈ R\I, the main formula allows us to compute the Hilbert-Poincaré series of R/(I, f) by means of the Hilbert-Poincaré series of R/I and R/(I : (f)), which are simpler to compute than HPR/(I,f ).
Let us compute Hilbert function and Hilbert-Poincaré series in the sim-plest case M = R with W = (1, . . . , 1) ∈ Nk.
8 CHAPTER 1. BASIC KNOWLEDGE Proposition 1.21 Let R = K[x1, . . . , xk]be a standard graded polynomial
ring. Then HR(i) = k+i−1n−1
for every i ∈ N and HPR(t) = (1−t)1 k
Proof (Hint). Given i ∈ N, the Hilbert function HR(i) counts the number of monomials in k variables of degree i, which is expressed by the binomial
k+i−1 n−1 . Then HPR(t) = X i≥0 HR(i)ti = X i≥0 k + i − 1 n − 1 ti and, since (1 − t)−k =P i≥0 k+i−1 n−1 ti, we are done.
Let us compute an actual Hilbert-Poincarè series for R graded by a generic matrix of positive type.
Example 1.22. Let R = K[x1, x2]be graded by W =
1 2 0 1
. Then we have R(i1,i2)6= 0 if and only if i2 ≥ 0 and i1− 2i2 ≥ 0. In these degrees we have dimK(R(i1,i2)) = 1. Therefore we obtain
HPR(t1, t2) = X i2≥0 X i1−2i2≥0 ti1 1t i2 2 = P i2≥0t 2i1 1 t i1 2 1 − t1 = 1 (1 − t1)(1 − t21t2)
This example suggests that there might be a simple formula for the Hilbert-Poincaré series of a polynomial ring. Next theorem delivers a very satisfactory answer.
Theorem 1.23 Let R = K[x1, . . . , xk]be graded by a positive matrix W =
(wij) ∈Matm,k(Z). Then HPR(t1, . . . , tm) = 1 Qk j=1(1 − t w1j 1 · · · t wmj m ) (1.4) Proof. We use induction on k, the number of variables of R.
For k = 1, we have degW(xi1) = (iw11, . . . , iwm1). Therefore we obtain
HPR(t1, . . . , tm) = X i≥0 tiw11 1 · · · t iwm1 m = 1 1 − tw11j· · · twmj m
and then the formula holds.
Now consider the case k > 1. Since degW(xk) = (w1k, . . . , wmk), then
Equa-tion (1.3) yields
HPR/(xk)(t) = (1 − t
w1k
1.4. HILBERT-SERRE THEOREM AND HILBERT POLYNOMIAL 9 Now we use the fact R/(xk) ∼= K[x1, . . . , xk−1]and the induction hypothesis
to finish the proof.
Equation (1.4) can be written concisely in the following way: HPR(t) = 1 Qk j=1(1 − tdegW(xj)) where tdegW(xj) = tw1j 1 · · · t wmj m .
1.4
Hilbert-Serre theorem and Hilbert polynomial
Let us focus definitively our attention on the case M = R/I, where I is a homogeneous ideal of R. Actually, in view of the next proposition we might consider only monomial ideals.
Before all, recall some preliminary notions.
We say that a total ordering σ on the terms of a ring R is a term-order, and we write σ ∈ TO(R), if t >σ 1 for all terms t of R.
We denote the ideal generated by the leading terms of polynomials in I by LTσ(I), namely LTσ(I) := (ltσ(f )|f ∈ I).
Proposition 1.24(Macaulay’s Lemma) Let I be a homogeneous ideal of a ring R and σ ∈ TO(R). Then
HR/I(d) = HR/LTσ(I)(d) for all d ∈ Z
m and HP
R/I = HPR/LTσ(I).
Proof. If we prove that dimK((R/I)d) = dimK((R/LTσ(i))d)for all d ∈ Zm,
then the first part of the statement follows.
For a graded piece (R/I)d, we have a vector space basis given by {f1, . . . , fj}.
We can assume that lt(f1) >σ · · · >σ lt(fj). If they don’t span (R/LTσ(I))d,
pick m ∈ (R/LTσ(I))d which is not in the span, such that lt(m) = m0 is
minimal with respect to the term order σ. Of necessity, there is a polynomial g ∈ R/I with lt(g) = m0. Since g is a K-linear combination of the fi, m0
must be one the lt(fi) (since no cancellation can occur), contradicting the
choice of m0 minimal.
The second part of the statement follows in a banal way by the first. We notice that in the proof of Macaulay’s lemma choice of term-order σ is irrelevant.
10 CHAPTER 1. BASIC KNOWLEDGE Next theorem gives us a characterisation of the Hilbert-Poincarè series HPR/I, in fact it shows that HPR/I is a rational function.
Theorem 1.25 (Hilbert-Serre) Let R := K[x1, . . . , xk] be a polynomial
ring graded by a matrix W = (wij) ∈Matm,k(Z) of positive type and let I
a homogeneous ideal of R. Then there exists a polynomial f ∈ Z[t1, . . . , tm]
such that HPR/I(t1, . . . , tm) = f (t1, . . . , tm) Qk j=1(1 − t w1j 1 · · · t wmj m )
Proof. By Macaulay’s lemma, we can suppose I is a monomial ideal and let {p1, . . . , ph} be a minimal monomial system of generators for I, i.e. I =
(p1, . . . , ph). We denote the ideal (p1, . . . , ph−1)by J and by the main formula
we get
HPR/I(t) = HPR/(J,ph)(t) = HPR/J− t
degW(ph)HP
R/(I:(ph))(t)
We observe that J and (I : (ph))have a minimal system of generators smaller
than that of I, so we can reduce the computation of the Hilbert-Poincaré series of R/I to that of two ideals with a less number of generators. We can interact this argument so that the resulting recursive algorithm to compute HPR/I finishes after a finitely number of steps and it yields a polynomial in tmultiplied by HPRwhich is equal to Qk 1
j=1(1−t w1j 1 ···t
wmj
m ), then the statement
follows.
The polynomial f(t1, . . . , tm)which appears in the denominator of HPR/I
is called h-vector and it is denote by < I >W or simply < I >.
In the proof of Hilbert-Serre theorem we have briefly illustrated a pro-cedure to compute the h-vector. Before applying it to a concrete case, we make a list of some useful rules which derive from the main formula.
Let xα= xα1 1 · · · x αk k be a term of R. Then (a) < 0 > = 1; (b) < 1 > = 0; (c) < xα> = 1 − tdegW(xα) ; (d) < xαI > = 1 − tdegW(xα)+ tdegW(xα) < I >; (e) < I, xα> = < I > −tdegW(xα)< I : (xα) >;
1.4. HILBERT-SERRE THEOREM AND HILBERT POLYNOMIAL 11 Example 1.26. Let R = [x, y, z] be graded by W = 1 1 2
0 3 1
, and let I = (x2y, y3, z2). We want to compute the Hilbert-Poincaré series of R/I following the algorithm sketched in the proof of Hilbert-Serre theorem. Let us consider I = (J, z2), where J = (x2y, y3). Then, by rule (e) we get
< I > = < J > −tdegW(z2)< J : (z2) >
Then we have to compute < J > and < J : (z2) >. Start from the
compu-tation of < J >. Let us consider J = (J0, y3), where J0 = (x2y) and again
by the rule (e) we get
< J > = < x2y > −tdegW(y3)< J0 : (y3) >
By rule (c), we have < x2y >= 1 − tdegW(x2y), and, since J0 : (y3) = (x2),
< J0 : (y3) >= 1 − tdegW(x2). Then
< J > = (1 − tdegW(x2y)) − tdegW(y3)(1 − tdegW(x2))
= (1 − t31t32) − t31t92(1 − t21)
Let us compute < J : (z2) >. We observe that < J : (z2) > = < J >, then
we have found < I > = (1 − tdegW(z2)) < J > = (1 − t4 1t22)[(1 − t31t32) − t31t92(1 − t21)] Finally, we have HPR/I = < I >
(1 − tdegW(x))(1 − tdegW(y))(1 − tdegW(z))
= −t
9
1t112 + t71t112 + t71t52+ t51t92− t41t22− t13t92− t31t32
(1 − t1)(1 − t1t32)(1 − t21t2)
Remark 1.27. By recalling that for all t = (t1, . . . , tm) and for all α ∈ Nm it
holds
(1 − tα)−1 =X
i≥0
(tα)i
it follows that the computation of HR/I from HPR/I is quite easy. In fact, it
suffices to invert the denominator and multiply the result by the numerator to get the canonical representation of the Hilbert-Poincaré series from which it is immediate to read the Hilbert function.
The main corollary of Hilbert-Serre theorem is the next result which shows that, under certain hypothesis, there exists a unique polynomial PR/I(n)
12 CHAPTER 1. BASIC KNOWLEDGE Corollary 1.28Let R be a polynomial standard graded ring and let I a homogeneous ideal of R. There exists a polynomial PR/I(t) ∈ Q[t] such that
HR/I(n) = PR/I(n)for all sufficiently large integer n 0. Proof. By Hilbert-Serre theorem, we have
HPR/I(t) =
f (t) (1 − t)r =
h(t) (1 − t)s
where h(t) ∈ Z[t] is a polynomial such that h(1) 6= 0 and s ≤ r. Let h(t) = N X n=0 antn and (1 − t)−s= ∞ X n=0 n + s − 1 s − 1 tn, then HPR/I(t) =X n≥0 " N X i=0 ai n + s − i − 1 s − 1 # tn
So, for n ≥ N, we have
HR/I(n) = N X i=0 ai n + s − i − 1 s − 1
which is a polynomial in n with leading term h(1) (s−1)!n
s−1 6= 0.
The polynomial PR/I(t)defined in the corollary 1.28 is called the Hilbert
polynomial of R/I.
Aim of next chapter is to show the existence of a sort of Hilbert poly-nomial for rings graded by a weight vector with entries not necessarily all equal to 1. Before treating this case, we make some remarks about Hilbert polynomial.
By the proof of corollary 1.28, we get that PR/I(t)is a polynomial with
rational coefficients and, if the Hilbert-Poincaré series is HPR/I(t) = h(t) (1 − t)s = a0+ a1t + . . . aNtN (1 − t)s with h(1) 6= 0, we have PR/I(t) = N X i=0 ai t + s − i − 1 s − 1
1.4. HILBERT-SERRE THEOREM AND HILBERT POLYNOMIAL 13 formula for PR/I(t). In addition, the degree of PR/I(t)is s−1 and its leading
coefficient is h(1) (s−1)!.
Example 1.29. Let R = K[x, y, z] and let I = (x2y, xy2, z3)be an ideal of R. The Hilbert-Poincaré series of R/I is
HPR/I(t) = 1 + 2t + 3t
2+ t3− t5
1 − t
Then, by using the formula above with s = 1, N = 5, a0 = 1, a1 = 2, a2 = 3,
a3= 1, a4 = 0 and a5 = −1, we get PR/I(t) = 5 X i=0 ai t + 1 − i − 1 1 − 1 = 5 X i=0 ai= 6
for all t > 5. Thus, we have
HR/I(n) = 1 if n = 1 3 if n = 2 5 if n = 3 6 if n = 4 5 if n = 5 6 if n ≥ 6
Chapter 2
Hilbert quasi-polynomial
In this chapter, we consider a polynomial ring R over an infinite field K of characteristic 0 where the degrees of the variables are assumed to be positive integers with no further restriction. We prove that the Hilbert function of R/Iis definitely equal to a quasi-polynomial PR/IW , where I is a homogeneous ideal of R, and we investigate the structure of PW
R/I, such as its degree, its
leading coefficient and proprieties of its coefficients.
2.1
Existence of the Hilbert quasi-polynomial
From now on, (R, W ) stands for a polynomial ring R := K[x1, . . . , xk]
with the graduation given by W := [d1, . . . , dk] ∈ Nk+. Let I be a
homoge-neous ideal of (R, W ). Thanks to Macaulay’s lemma, we can suppose that I is a monomial ideal. In addition, we denote the least common multiple of d1, . . . , dk by d. We are going to prove that the Hilbert function of (R/I, W )
is definitely equal to a quasi-polynomial of period d.
Definition 2.1 A function f : N → N is a quasi-polynomial of period s if there exist polynomials p0, . . . , ps−1 ∈ Q[x] such that f(n) = pi(n) when
n ≡ i mod s.
Proposition 2.2 Let (R, W ) and I be as above, there exists a uniquely determined quasi-polynomial PW
R/I of period d such that HR/I(n) = P W R/I(n)
for all n 0.
Proof. From the Hilbert-Serre theorem, we know that the Hilbert-Poincaré series of R/I can be written as a rational function
HPR/I(t) = h(t) Qk i=1(1 − tdi) ∈ Q[|t|] 15
16 CHAPTER 2. HILBERT QUASI-POLYNOMIAL Let ζ be a primitive dth root of unity, so we can write
k Y i=1 (1 − tdi) = d−1 Y j=0 (1 − ζjt)αj, for some α j ∈ N such that d−1 X j=0 αj = k X i=1 di
By using this relation and partial fractions, we have HPR/I(t) = h(t) Qd−1 j=0(1 − ζjt)αj = d−1 X j=0 Qj(t) (1 − ζjt)αj (2.1)
where Qj(t) ∈ C[t] is a polynomial of degree nj for all j = 0, . . . , d − 1,
namely Qj(t) =Pnh=0j ajhth.
We are going to consider the addends individually, so Qj(t) (1 − ζjt)αj = nj X h=0 ajhth ! · 1 (1 − ζjt)αj = nj X h=0 ajhth ! · X n≥0 n + αj − 1 n ζjntn =X n≥0 "nj X h=0 ajh n + αj− h − 1 n − h ζj(n−h) # tn =X n≥0 "nj X h=0 ajh n + αj− h − 1 αj− 1 ζj(n−h) # tn=X n≥0 bjntn where bjn:= nj X h=0 ajh n + αj− h − 1 αj− 1 ζj(n−h)
We can rewrite the Hilbert-Poincaré series in the following way: HPR/I(t) = d−1 X j=0 Qj(t) (1 − ζjt)αj = d−1 X j=0 X n≥0 bjntn= X n≥0 d−1 X j=0 bjn tn
and, therefore, for all n ≥ max
j {nj}, we have HR/I(n) = d−1 X j=0 bjn= d−1 X j=0 nj X h=0 ajh n + αj− h − 1 n − h ζj(n−h)= d−1 X j=0 Sj(n)ζjn where Sj(n) := nj X h=0 ajh n + αj− h − 1 n − h ζ−jh.
2.1. EXISTENCE OF THE HILBERT QUASI-POLYNOMIAL 17 So, given n ≥ max
j {nj}, let i be an integer such that 0 ≤ i ≤ d − 1 and
i ≡ n mod d, the Hilbert function evaluated at n is equal to the evaluation at n of the polynomial
Pi(x) := S0(x) + S1(x)ζi+ · · · + Sd−1(x)ζ(d−1)i.
The quasi-polynomial PW
R/I = {P0, . . . , Pd−1} defined in proposition 2.2 is
called the Hilbert quasi-polynomial of (R/I,W).
By the proof, we find out that HR/I(n) = PR/IW (n)for all n ≥ max j {nj},
where nj are the degrees of the polynomials Qj(t)which appear in Equation
(2.1) of the Hilbert-Poincaré series.
Remark 2.3. Proposition 2.2 asserts that the Hilbert quasi-polynomial has period equal to d, but it doesn’t assure that the d polynomials are necessarily all distinct. We’ll see in next examples that it can happen.
In addition, in the proof of proposition we yield a way to compute the Hilbert quasi-polynomial of (R/I, W ), when the Hilbert-Poincaré series is known.
Recall that, thanks to the corollary 1.28, if R is a polynomial ring with standard graduation and I is a homogeneous ideal of R, then the Hilbert polynomial PR/I is a polynomial with rational coefficients and PR/I(n) ∈ N
for all sufficiently large n ∈ N. We’ll show that the Hilbert quasi-polynomial of (R/I, W ) satisfies the same proprieties. Before all, we prove a general result concerning polynomials and their coefficients.
Theorem 2.4 Let f(x) ∈ C[x] be a polynomial of degree n. If f(xi) ∈ Z
for at least n + 1 points xi ∈ Z, then f(x) ∈ Q[x].
Proof. Consider f(x) = anxn+ · · · + a1x + a0. By hypothesis, there exist
x0, . . . , xn ∈ Z such that f(xi) = yi ∈ Z. If we substitute the equation of
f (x)in here, we get a system of linear equations in the coefficients ak. The
system in matrix-vector form reads xn0 xn−10 xn−20 . . . x0 1 xn1 xn−11 xn−21 . . . x1 1 ... ... ... ... ... xnn xn−1n xn−2n . . . xn 1 an an−1 ... a0 = y0 y1 ... yn
Let V be the matrix on the left and let b be the vector [y0, y2, . . . , yn]T.
18 CHAPTER 2. HILBERT QUASI-POLYNOMIAL f (x)are uniquely determined by
[an, an−1, . . . , a0]T = V−1b
Since b ∈ Zn+1 and V ∈ Mat
n+1(Z), it holds V−1 ∈ Matn+1(Q) and the
statement follows.
Proposition 2.5Let (R, W ) and I be as above and let PR/IW = {P0, . . . , Pd−1}
be the Hilbert quasi-polynomial of (R/I, W ). Then Pi(x) ∈ Q[x] for all
i = 0, . . . , d − 1.
Proof. By definition, Pi(n) ∈ N for all sufficiently large n ∈ N such that
i ≡ n mod d, then the hypothesis of the theorem 2.4 are satisfied and we can conclude that PW
R/I is a quasi-polynomial with rational coefficient.
Thanks to the following propositions, we can restrict in a particular case: (R/I, W ) with I = (0) and W = [d1, . . . , dk] a weight vector such that
gcd(d1, . . . , dk) = 1. In fact, for any homogeneous ideal I of R we can
com-pute the Hilbert quasi-polynomial PW
R/Istarting from PRW, and given a weight
vector W0 := [d0
1, . . . , d0k]such that c := gcd(d01, . . . , d0k) 6= 1, we can compute
the Hilbert quasi-polynomial PW0
R starting from PRW, with W := 1 cW
0.
Proposition 2.6Let PRW be the Hilbert quasi-polynomial of (R, W ). Let I be a homogeneous ideal of R and let HPR/I(t) = Qk h(t)
i=1(1−tdi)
be the Hilbert-Poincaré series of R/I, where h(t) := Pr
j=0ajtj. Then, for all n sufficiently
large, the Hilbert quasi-polynomial PW
R/I is given by PR/IW (n) = r X j=0 ajPRW(n − j) Proof. Since HPR(t) = X n≥0 HR(n)tn= 1 Qk i=1(1 − tdi) we have HPR/I(t) = h(t) Qk i=1(1 − tdi) = X n≥0 HR(n)tn r X j=0 ajtj =X n≥0 r X j=0 ajHR(n − j) tn
2.1. EXISTENCE OF THE HILBERT QUASI-POLYNOMIAL 19 with the convention HR(n) = 0 for all n < 0. Therefore, for all n ≥ r,
HR/I(n) =
Pr
j=0ajHR(n − j) . Given that HR(n) = PRW(n) for all
suffi-ciently large n, then HR(n − j) = PRW(n − j) for all sufficiently large n and
so HR/I(n) =
Pr
j=0ajPRW(n − j) for all sufficiently large n. The statement
follows.
Proposition 2.7Let PRW = {P0, . . . , Pd−1}be the Hilbert quasi-polynomial
of (R, W ). If W0 := a · W = [d0
1, . . . , d0k] for some a ∈ N+, then the Hilbert
quasi-polynomial PW0
R = {P00, . . . , Pad−10 } of (R, W0) is such that:
(i) If a doesn’t divide i, then P0 i = 0 ;
(ii) If a divides i, then P0
i(x) = Pi a(
x a) .
Proof.
(i) Let i be a positive integer such that a doesn’t divide i. We observe that HRW0(n) = 0for all n ∈ N such that n ≡ i mod a. Indeed, assume that HRW0(n) 6= 0 for some n ≡ i mod a. Then there exist a1, . . . , ak ∈ N
such that a1d01+ · · · + akd0k = n. By hypothesis, a divides d 0
1, . . . , d0k,
hence a divides n and so a divides i, which is a contradiction. Since Pi0(n) = HRW0(n)for all sufficiently large n such that n ≡ i mod a, then Pi0 has an infinitive number of roots and so Pi0= 0.
(ii) Recall that for all sufficiently large n such that n ≡ i mod a, we have Pi(n) = HRW(n) = #{(a1, . . . , ak) ∈ Nk|a1d1+ . . . akdk= n} Then, Pi0(n) = HRW0(n) = #{(b1, . . . , bk) ∈ Nk | ab1d1+ · · · + abkdk= n} = #{(b1, . . . , bk) ∈ Nk | b1d1+ · · · + bkdk= n a} = HRW(n a) = Pai( n a)
To avoid endlessly repeating these hypothesis, we shall use W to denote a weight vector [d1, . . . , dk] such that gcd(d1, . . . , dk) = 1, throughout the
20 CHAPTER 2. HILBERT QUASI-POLYNOMIAL
2.2
Degree and leading coefficient of the Hilbert
quasi-polynomial
In this section, we investigate general proprieties of the Hilbert quasi-polynomial of (R, W ).
First of all, we introduce the following notation which will help us to rewrite the denominator of the Hilbert-Poincaré series of (R, W ) in a suit-able way.
Given d1, . . . , dk∈ N+, with d := lcm(d1, . . . , dk), we define
• δ := max{|I| | gcd(di)i∈I 6= 1and I ⊆ {1, . . . , k}},
• ˆds:= dds, • Ms := { ˆds, 2 ˆds, . . . , (ds− 1) ˆds} for all s = 1, . . . , k, • Tr:= [ J ⊆{1,...,k} |J|=r (\ s∈J Ms) for all r = 1, . . . , k.
Observe that Tr is the set of the elements p ∈ Ski=1Mi which belong to
exactly r sets Mi.
Now we’ll analyse the denominator of the Hilbert-Poincaré series of (R, W ), which we denote by g(t). Let ζ := ζd be a primitive dth root of unity. We
remark that ζdj = ζ
ˆ
dj for all j = 1, . . . , k. Then we can write
g(t) = k Y i=1 (1 − tdi) = (1 − t)k k Y i=1 h (1 − ζdˆit) · · · (1 − ζ(di−1) ˆdit) i = (1 − t)k Y j∈M1 (1 − ζjt) · · · Y j∈Mk (1 − ζjt) = (1 − t)k Y j∈Tk (1 − ζjt)k· · · Y j∈T1 (1 − ζjt) (2.2)
Example 2.8. Let W = [2, 3, 4]. By using the notation introduced above, we have d = 12, δ = 2, ˆd1 = 6, ˆd2= 4 and ˆd3 = 3.
2.2. DEGREE AND LEADING COEFFICIENT 21 T2 = {6}and T3= ∅.
Let ζ := ζ12, then we can rewrite g(t) in the following way:
g(t) = (1 − t2)(1 − t3)(1 − t4)
= (1 − t)3(1 − ζ6t) (1 − ζ4t)(1 − ζ8t) (1 − ζ3t)(1 − ζ6t)(1 − ζ9t) = (1 − t)3(1 − ζ6t)2 (1 − ζ3t)(1 − ζ4t)(1 − ζ8t)(1 − ζ9t)
We are now going to show some lemmas concerning the sets Ms and Tr.
Lemma 2.9Let d1, . . . , dk ∈ N+. For any subset {j1, . . . , jr} ⊆ {1, . . . , k},
it holds Mj1 ∩ · · · ∩ Mjr = d gcd(dj1, . . . , djr) , 2 · d gcd(dj1, . . . , djr) , . . . . . . , (gcd(dj1, . . . , djr) − 1) · d gcd(dj1, . . . , djr)
Proof. We use induction on r, the number of intersected subsets.
The case r = 1 is trivial. Suppose r ≥ 2 and consider, without loss of generality, the subsets M1, . . . , Mr instead of Mj1, . . . , Mjr. By induction
hypothesis, we have M1∩ · · · ∩ Mr−1 = d gcd(d1, . . . , dr−1) , 2 · d gcd(d1, . . . , dr−1) , . . . . . . , (gcd(d1, . . . , dr−1) − 1) · d gcd(d1, . . . , dr−1)
We observe that M1∩ · · · ∩ Mr−1 has the same structure of Ma, with a :=
gcd(d1, . . . , dr−1)(we can suppose that a is one of the given integers, without
undermining the statement). Therefore M1∩ · · · ∩ Mr = Ma∩ Mr and, by
using again the induction hypothesis, we obtain M1∩ · · · ∩ Mr = d gcd(a, dr) , 2 · d gcd(a, dr) , . . . , (gcd(a, dr) − 1) · d gcd(a, dr)
and since a = gcd(d1, . . . , dr−1), we are done.
22 CHAPTER 2. HILBERT QUASI-POLYNOMIAL Proof. We assume that Th 6= ∅ for some h = δ + 1, . . . , k. Consider p ∈ Th
and suppose, without loss of generality, that p ∈ M1∩ · · · ∩ Mh. By lemma
2.9, p = p1·gcd(d1d,...,dh) with 1 ≤ p1 ≤ gcd(d1, . . . , dh) − 1. Since δ < h, we
must have gcd(d1, . . . , dh) = 1. Therefore p is a multiple of d, which is a
contradiction because of p < d.
By lemma 2.10, in Equation (2.2) the set Tδ+1, . . . , Tk are empty, so
ultimately we have g(t) = k Y i=1 (1 − tdi) = (1 − t)k Y j∈Tδ (1 − ζjt)δ · · · Y j∈T1 (1 − ζjt) (2.3) and we obtain the following useful expression for the Hilbert-Poincaré series HPR(t) = 1 g(t) = Q0(t) (1 − t)k + X j∈Tδ Q(δ)j (t) (1 − ζjt)δ + · · · + X j∈T1 Q(1)j (t) (1 − ζjt) (2.4)
Remark 2.11. We have already seen in the proof of proposition 2.2 that an addend Qj(t)
(1−ζjt)αj of HPR(t), with Qj(t) =Pnh=0j ajhth, can be written in the
following way Qj(t) (1 − ζjt)αj = X n≥0 "nj X h=0 ajh n + αj− h − 1 αj− 1 ζj(n−h) # tn (2.5) Then the rational function Qj(t)
(1−ζjt)αj corresponds to a power series in t where
the nth coefficient depends on n in a polynomial way through the binomial
n+αj−h−1
αj−1
and its degree in n is αj− 1.
We consider the equation (2.4) of the Hilbert-Poincaré series and, by using partial fractions, we get
HPR(t) = 1 g(t) = A1 1 − t + A2 (1 − t)2 + · · · + Ak (1 − t)k+ + X j∈Tδ Bj,1(δ) 1 − ζjt + · · · + B(δ)j,δ (1 − ζjt)δ + · · · + X j∈T1 " Bj,1(1) 1 − ζjt # (2.6)
2.2. DEGREE AND LEADING COEFFICIENT 23 for some Ai, B(r)j,h ∈ Q.
Remark 2.12. By multiplying both sides of Equation (2.6) by g(t), we get
1 = k X t=1 At(1 − t)k−t· Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) + + X j∈Tδ (1 − t)k· δ−1 Y i=1 Y r∈Ti (1 − ζjt)i · k X m=1 Bj,m(δ) · Y s∈Tδ s6=j (1 − ζst)δ−m + + ... + + X j∈T1 Bj,1(1)(1 − t)k· Y r∈Tδ (1 − ζrt)δ· · · Y r∈T2 (1 − ζr)2 Y s∈T1 s6=j (1 − ζst) (2.7)
Remark 2.13. By Equation (2.6), we have
Q0(t) = A1(1 − t)k−1+ A2(1 − t)k−2+ · · · + Ak (2.8)
Lemma 2.14 The constant Ak which appears in Equation (2.6) and (2.8) is such that Ak= 1 Qk i=1di .
Proof. By evaluating Equation (2.7) at t = 1, we get 1 = Ak· Y j∈Tδ (1 − ζj)δ· · · Y j∈T1 (1 − ζj) If we prove that Qj∈δ(1 − ζj)δ· · · Q j∈T1(1 − ζ j) is equal to Qk i=1di, we are
24 CHAPTER 2. HILBERT QUASI-POLYNOMIAL which derives from equation (2.3):
g(t) (1 − t)k = Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) g(t) (1 − t)k = k Y j=1 1 − tdj 1 − t = k Y j=1 (1 + t + · · · + tdj−1) Then, it holds Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) = k Y j=1 (1 + t + · · · + tdj−1) (2.9)
By evaluating both sides of Equation (2.9) at t = 1, the statement follows.
Proposition 2.15Let (R, W ) and PRW be as above. The degree of Pi is
equal to k − 1 for all i = 0, . . . , d − 1 and its leading coefficient lc(Pi)is such
that
lc(Pi) =
1 (k − 1)!Qk
i=1di
Proof. Since δ < k by hypothesis and thanks to Equation (2.4) and remark 2.11, it suffices to analyse the term
Q0(t) (1 − t)k = X n≥0 "k−1 X h=0 ah n − h + k − 1 k − 1 # tn
where ah:= a0,h for all h = 0, . . . , k − 1.
We observe that Pk−1
h=0ah n−h+k−1k−1 = S0(n)and then the degree of Pi
is less or equal to k − 1. It remains to show that the coefficient of nk−1 is
not 0. In particular, we wish to show that it is equal to 1 (k−1)!Qk
i=1di.
Suppose n ≥ k − 1, then the nth coefficient of HPR(t)is equal to
S0(n) = k−1 X h=0 ah n − h + k − 1 k − 1 = k−1 X h=0 ah (n − h + k − 1)(n − h + k − 2) · · · (n − h + 1) (k − 1)! = k−1 X h=0 ah· nk−1
2.2. DEGREE AND LEADING COEFFICIENT 25 Then the leading coefficient of S0(n)is
1 (k − 1)! k−1 X h=0 ah = Q0(1) (k − 1)! = Ak (k − 1)! = 1 (k − 1)!Qk i=1di as required.
Now, we’ll give a degree bound for the Hilbert quasi-polynomial of (R/I, W ), when I 6= (0).
Proposition 2.16Let I 6= (0) a homogeneous ideal of R and let PR/IW the Hilbert quasi-polynomial of (R/I, W ). Then the degree of Pi is less or equal
to k − 2.
In order to prove Proposition 2.16, we need the following results.
Let I 6= (0) be a homogeneous monomial ideal and consider the Hilbert-Poincaré series HPR/I(t) = Qk f (t)
i=1(1−tdi)
= f (t)g(t) of (R/I, W ).
Let f(t) := q(t)g(t) + r(t), where q(t), r(t) ∈ Q(t) and deg r < deg g. Then, we can write
HPR/I(t) = f (t) g(t) = q(t) + r(t) g(t) = q(t) + C1 1 − t+ C2 (1 − t)2 + · · · + Ck (1 − t)k+ +X j∈Tδ Dj,1(δ) 1 − ζjt+ · · · + D(δ)j,δ (1 − ζjt)δ + · · · + X j∈T1 " Dj,1(1) 1 − ζjt # (2.10) for some Ci, Dj,h(r)∈ Q.
Lemma 2.17 The constant Ck which appears in Equation (2.10) is zero.
26 CHAPTER 2. HILBERT QUASI-POLYNOMIAL r(t) = k X t=1 Ct(1 − t)k−t· Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) + +X j∈Tδ (1 − t)k· δ−1 Y i=1 Y r∈Ti (1 − ζjt)i · k X m=1 Dj,m(δ) · Y s∈Tδ s6=j (1 − ζst)δ−m + + ... + +X j∈T1 Dj,1(1)(1 − t)k· Y r∈Tδ (1 − ζrt)δ· · · Y r∈T2 (1 − ζr)2 Y s∈T1 s6=j (1 − ζst) (2.11) By evaluating Equation (2.11) at t = 1, we have
r(1) = Ak· Y j∈T δ (1 − ζj)δ· · · Y j∈T1 (1 − ζj)
We already know that Qj∈T δ(1 − ζj)δ· · ·Q
j∈T1(1 − ζ
j) =Qk
i=1di, then we
have to show that r(1) = 0. Actually, it suffices to prove that f(1) = 0, because f(1) = q(1)g(1) + r(1) and f(1) = g(1) = 0 imply r(1) = 0.
We prove that f(1) = 0 by induction on the cardinality s of a minimal set of generators of I.
Suppose s = 1, since I 6= (0) there exists m ∈ I such that I = (m). Then f (t) =< I >= 1 − tdegW(m), so obviously f(1) = 0.
If s ≥ 2, let {m1, . . . , ms} a minimal set of generators of I, then I =
(m1, . . . , ms) and, by using the main formula, we have
f (t) = < (m1, . . . , ms) > =
< (m1, . . . , ms−1) > −tdegW(ms)< (m1, . . . , ms−1) : (ms) >
and we can conclude by induction hypothesis.
We are now able to demonstrate that for all i = 0, . . . , d − 1 it holds deg(Pi) ≤ k − 2.
2.2. DEGREE AND LEADING COEFFICIENT 27 Proof of Proposition 2.16. We consider the following expression for the Hilbert-Poincaré series, which we have already seen in Equation (2.10)
HPR/I(t) = f (t) g(t) = q(t) + r(t) g(t) = q(t) + C1 1 − t+ C2 (1 − t)2 + · · · + Ck (1 − t)k+ +X j∈Tδ Dj,1(δ) 1 − ζjt+ · · · + D(δ)j,δ (1 − ζjt)δ + · · · + X j∈T1 " Dj,1(1) 1 − ζjt # Since δ < k, and by lemma 2.17, which asserts that Ck= 0, we have that the
maximum exponent in the denominator of HPR/I is at most k−1. Therefore,
by recalling the construction of Hilbert quasi-polynomial done in the proof of proposition 2.2, we write HPR/I(t) = d−1 X j=0 Qj(t) (1 − ζjt)αj
for some polynomials Qj(t) := P nj
h=0ajhth ∈ C[t] of degree nj and with
αj ≤ k − 1 for all j = 0, . . . , d − 1. Then, for all n ≥ maxj{nj}, we have
HR/I(n) = d−1 X j=0 Sj(n)ζjn where Sj(n) := nj X h=0 n + αj − h − i αj− 1 ζ−jh
Observe that Sj is a polynomial in n of degree αj − 1.
Finally, we get Pi(x) := S0(x) + S1(x)ζi+ · · · + Sd−1(x)ζ(d−1)i and, by our
previous considerations, every Sj has degree less or equal to k − 2 and the
statement follows.
Example 2.18. Consider the ring R = Q[x1, . . . , x4] with the graduation
given by W = [1, 2, 4, 8]. Let I = (x2
1x3− x32, x4− x41x3) ⊂ R. In this case,
we have k = 4 and d = 8.
The Hilbert quasi-polynomial PW
R = {P0, . . . , P7} is given by P0(x) = 1/384x3+ 1/16x2+ 11/24x + 1 P1(x) = 1/384x3+ 7/128x2+ 131/384x + 77/128 P2(x) = 1/384x3+ 1/16x2+ 41/96x + 7/8 P3(x) = 1/384x3+ 7/128x2+ 119/384x + 65/128 P4(x) = 1/384x3+ 1/16x2+ 11/24x + 1 P5(x) = 1/384x3+ 7/128x2+ 131/384x + 77/128 P6(x) = 1/384x3+ 1/16x2+ 41/96x + 5/8 P7(x) = 1/384x3+ 7/128x2+ 119/384x + 33/128
28 CHAPTER 2. HILBERT QUASI-POLYNOMIAL For all i = 0, . . . , 7, the polynomials Pi have degree equal to k − 1 = 3 and
the leading coefficient is 1 (k−1)!Qk
i=1di =
1 384.
Whereas the Hilbert quasi-polynomial PW
R/I = {P0, . . . , P7} of (R/I, W ) is given by P0(x) = 7/4x − 5 P1(x) = 7/4x − 27/4 P2(x) = 7/4x − 11/2 P3(x) = 7/4x − 29/4 P4(x) = 7/4x − 5 P5(x) = 7/4x − 27/4 P6(x) = 7/4x − 11/2 P7(x) = 7/4x − 29/4
and we observe that the degree of P is 1, which is less than 2 = k − 2.
2.3
Proprieties of the coefficients of the Hilbert
quasi-polynomial
In this section we present some proprieties of the coefficients of Hilbert quasi-polynomials. In particular, if we denote the Hilbert quasi-polynomial of R by PW
R = {P0, . . . , Pd−1} as usual, we show that some coefficients are
equal for all Pi, while the others change with a certain regularity, as
sug-gested by examples that we have studied. These facts could be exploited in order to reduce complexity in the computation of Hilbert quasi-polynomials.
Before showing our results, we present an example.
Example 2.19. Consider R = Q[x1, . . . , x5] with the graduation given by
W = [1, 2, 3, 4, 6]. In this case, we have k = 5 and d = 12. The Hilbert quasi-polynomial PW
2.3. PROPRIETIES OF THE COEFFICIENTS 29 P0(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 1/2x + 1 P1(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 43/108x + 1705/3456 P2(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 25/54x + 125/216 P3(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 5/12x + 75/128 P4(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 13/27x + 20/27 P5(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 41/108x + 1001/3456 P6(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 1/2x + 7/8 P7(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 43/108x + 1705/3456 P8(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 25/54x + 19/27 P9(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 5/12x + 75/128 P10(x) = 1/3456x4+ 1/108x3+ 5/48x2+ 13/27x + 133/216 P11(x) = 1/3456x4+ 1/108x3+ 19/192x2+ 41/108x + 1001/3456
We have 12 polynomials of degree 4 with leading coefficient equal to 1/3456, that’s just what we expected.
As can be seen by comparing polynomials, the following facts hold:
• The coefficient of the term of degree 3 is the same for all polynomials. • The coefficient of the term of degree 2 has periodicity 2, i.e. 5/48 is the coefficient of the term of degree 2 of all Pi(x) such that i ≡ 0 mod 2
and 19/192 is the coefficient of the term of degree 2 of all Pi(x) such
that i ≡ 1 mod 2.
• The coefficient of the term of degree 1 has periodicity 6, i.e. 1/12 is the coefficient of the term of degree 1 of all Pi(x) such that i ≡ 0 mod 6,
43/108 is the coefficient of the term of degree 1 of all Pi(x) such that
i ≡ 1 mod 6, 5/48 is the coefficient of the term of degree 1 of all Pi(x)
such that i ≡ 2 mod 6 and so on.
• The constant term seems not to have any kind of regularity.
Goal of next two results is to see if we can predict which and how coeffi-cients change.
Recall that we use δ to denote max{|I| | gcd(di)i∈I 6= 1and I ⊆ {1, . . . , k}}.
Proposition 2.20 Let (R, W ) and PRW be as above. Then PRW(z) = Q(z) + R(z)
where Q(z) ∈ Q[z] is a polynomial of degree k − 1, whereas R(z) is a quasi-polynomial with rational coefficients of degree δ − 1.
30 CHAPTER 2. HILBERT QUASI-POLYNOMIAL Proof. Lemma 2.10 assures that Tδ+1 = · · · = Tk = ∅. Thus the
Hilbert-Poincaré series of (R, W ) is HPR(t) = Q0(t) (1 − t)k + X j∈Tδ Q(δ)j (t) (1 − ζjt)δ + · · · + X j∈T1 Q(1)j (t) (1 − ζjt) (2.12)
and, thanks to remark 2.11, we have that the coefficients of the terms of degrees δ, . . . , k − 1 don’t change and the result follows.
Let us return to the example 2.19. In that case, δ is equal to 3, in fact the subset {2, 4, 6} is the largest subset such that its elements are not relatively prime. By proposition 2.20, the coefficients of the terms of degree δ, . . . , k−1, that is 3 and 4, don’t change and this is exactly what happens in the example. Proposition 2.21Let (R, W ) and PRW be as above. The rth coefficient of PRW, for r = 0, . . . , k − 1, has periodicity equal to
δr:= lcm ( gcd(di)i∈I | |I| = r + 1, I ⊆ {1, . . . , k})
Formally, the proposition asserts that if we denote the rth coefficient of Pi(t)by air, then ajr = air when j = i + δr mod d.
Proof. Let δ = max{|I| | gcd(di)i∈I 6= 1 and I ⊆ {1, . . . , k}} be as usual.
Then δr = 1for δ ≤ r ≤ k −1 and, by proposition 2.20, the assertion follows.
Let r ≤ δ − 1. Recall that Pi(x) = S0(x) + S1(x)ζi+ · · · + Sd−1(x)ζ(d−1)i,
where Sj(x) =P nj
h=0ajh
x+αj−h−1
αj−1 ζ
−jh. Note that each S
j(x) doesn’t
de-pend on i and its degree is equal to αj− 1. Therefore the rth coefficient of
Pi(x) is the rth coefficient of S0(x) + Sj1(x)ζ i·j1 + · · · + S jv(x)ζ i·jv with {j1, . . . , jv} = [ s≥r+1 Ts = [ r+1≤s≤δ Ts (2.13)
where the second equality holds because of Tδ+1 = · · · = Tk = ∅.
We recall that Ts= [ I⊆{1,...,k},|I|=s gcd(di)i∈I6=1 d gcd(di)i∈I , . . . , (gcd(di)i∈I− 1) · d gcd(di)i∈I
To prove that the rth coefficient of S0(x) + Sj1(x)ζ
i·j1 + · · · + S
jv(x)ζ
i·jv
2.3. PROPRIETIES OF THE COEFFICIENTS 31 Equation (2.13).
Let N = 1, then r = δ − 1. We observe that ζδr·jt = 1for all t = 1, . . . , v, (in
fact, jtcan be written as a·gcd(ddi)i∈I for some integer 1 ≤ a ≤ (gcd(di)i∈I−1),
with |I| = r + 1; since gcd(di)i∈I divides δr, then d divides δr· jt). So we
have S0(x) + Sj1(x)ζ i·j1+ · · · + S ju(x)ζ i·jv = S0(x) + Sj1(x)ζ (i+δr)·j1+ · · · + S jv(x)ζ (i+δr)·jv
and we are done.
Let N ≥ 2, then r ≤ δ − 2, and let us suppose that the thesis is true for S
r0≤s≤δTs, with r0 > δ − N, which means that the r0th coefficient has period
equal to δr0.
We can order the set {j1, . . . , ju, ju+1, . . . , jv}in such a way that
• j1, . . . , ju∈ Ts, with s > r + 1, and ji∈ T/ r+1, for all i = 1, . . . , u;
• ju+1, . . . , jv ∈ Tr+1.
Let γ := lcm(gcd(di)i∈I | |I| > r + 1), then by induction hypothesis we have
S0(x) + Sj1(x)ζ i·j1+ · · · + S ju(x)ζ i·ju = S0(x) + Sj1(x)ζ (i+γ)·j1 + · · · + S ju(x)ζ (i+γ)·ju
Since γ divides δr, it follows
S0(x) + Sj1(x)ζ i·j1+ · · · + S ju(x)ζ i·ju = S0(x) + Sj1(x)ζ (i+δr)·j1 + · · · + S ju(x)ζ (i+δr)·ju
In addition, since ζδr·jt = 1 for all t = u + 1, . . . , v, we have
Sju+1(x)ζ i·ju+1 + · · · + S jv(x)ζ i·jv = Sju+1(x)ζ (i+δr)·ju+1+ · · · + S jv(x)ζ (i+δr)·jv
and then, by putting together the two equations above, we are done.
Remark 2.22. In some particular cases, for example when k >> δ or if the periods δrare quite small, knowing the period of the coefficients could be
ex-ploited to reduce the cost of computation for the Hilbert quasi-polynomials. Return to example 2.19. We already know that δ = 3, then δr = 1 for
32 CHAPTER 2. HILBERT QUASI-POLYNOMIAL For r = 2, δr is equal to 2, in fact {2, 4, 6} is the only subset of cardinality
3 such that its elements are not relatively prime and gcd(2, 4, 6) = 2. For r = 1, δr is equal to 6, in fact the subsets of cardinality 2 such that
their elements are not relatively prime are {2, 4}, {2, 6}, {3, 6} and {4, 6} and their great common divisor are respectively 2, 2, 3 and 2. Therefore, δr= lcm(2, 2, 3, 2) = 6.
2.4
Formulas for the other coefficients of the Hilbert
quasi-polynomial
In this section, we give formulas to recover the (k − 2)th and (k − 3)th coefficient of the Hilbert quasi-polynomial from the weights d1, . . . , dk, and
we illustrate how recover other coefficients.
We have already showed that if δ ≤ k − 2 (δ ≤ k − 3 respectively) then the (k − 2)th ((k − 3)th respectively) coefficient is the same for all Pi,
i = 0, . . . , d − 1. For this reason, we suppose that δ is small enough to guar-antee that the coefficient is equal for all Pi.
Proposition 2.23Let (R, W ) and PRW as above. If δ ≤ k − 2, we denote the (k − 2)th coefficient of Pi, for all i = 0, . . . , d − 1, by ck−2. Then, it holds
ck−2=
Pk
i=1di
2 · (k − 2)! ·Qk
i=1di
Proof. Let Pi(n) = S0(n) + S1(n)ζn+ · · · + Sd−1(n)ζn·(d−1) as usual, where
Sj(n) = nj X h=0 ajh n + αj− h − 1 αj− 1 ζ−jh.
Since δ ≤ k − 2 and by remark 2.11, the (k − 2)th coefficient of Pi is given
by the (k − 2)th coefficient of S0(n). Let ai:= a0i, then we have
S0(n) = a0 n + k − 1 k − 1 +a1 n + k − 2 k − 1 +· · ·+ak−1 n k − 1 +ak n − 1 k − 1 =
2.4. FORMULAS FOR THE OTHER COEFFICIENTS 33 = a0· (n + k − 1) · · · (n + 1) (k − 1)! + a1· (n + k − 2) · · · (n) (k − 1)! + + · · · + ak−1· (n) · · · (n − k + 2) (k − 1)! + ak· (n − 1) · · · (n − k + 1) (k − 1)! = k−1 X i=0 cini (2.14) By this equation, we get
ck−2= a0 (k − 1) + (k − 2) + · · · + 1 (k − 1)! + a1 (k − 2) + (k − 3) + · · · + 1 (k − 1)! + + · · · + ak (−1) + (−2) + · · · + (−k + 1) (k − 1)! = = a0 (k − 1)!· k(k − 1) 2 + a1 (k − 1)!· (k − 1)(k − 2) 2 +· · ·+ ak (k − 1)!· (−k)(k − 1) 2 = = 1
2(k − 2)![ka0+ (k − 2)a1+ · · · + (k − 2k)ak] =
= 1 2(k − 2)! " k k X i=0 ai− 2 k X i=0 (i · ai) # (2.15) We observe that k X i=0 ai= Q0(1) and k X i=0 i · ai = Q00(1) (2.16) Since Q0(t) = A1(1 − t)k−1+ A2(1 − t)k−2 + · · · + Ak−1(1 − t) + Ak, by
evaluating Q0(t)and Q00(t)at t = 1 and by putting together Equation (2.15)
and Equation (2.16), we have ck−2= 1 2(k − 2)!k · Q0(1) − 2 · Q 0 0(1) = 1 2(k − 2)![k · Ak+ 2 · Ak−1] . By the following lemma and by recalling that Ak = Qk1
i=1di, the proposition
is proved.
Lemma 2.24 The constant Ak−1 which appears in Equation (2.6) of the
expression of Q0(t)is such that
Ak−1 =
Pk
i=1di− k
2 ·Qk
34 CHAPTER 2. HILBERT QUASI-POLYNOMIAL
Proof. In order to obtain an expression for Ak−1, evaluating at t = 1 the derivative of the Equation (2.7) is a good strategy. Before doing it, we observe that the evaluation at t = 1 of the derivative of the equation gives 0 except for the terms in the first summation for t = k − 1 and t = k. For this reason, we’ll consider only these addends and we’ll calculate only their derivatives. 0 = ∂ ∂t (Ak−1(1 − t) + Ak) · Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 0 = −Ak−1· Y j∈Tδ (1 − ζj)δ· · · Y j∈T1 (1 − ζj)+ + Ak· ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1
We already know that Ak = 1 Qk i=1di and Y j∈Tδ (1 − ζj)δ· · · Y j∈T1 (1 − ζj) = k Y i=1 di. So, we have Ak−1 = 1 (Qk i=1di)2 · ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 (2.17)
and it remains to calculate ∂ ∂t h Q j∈Tδ(1 − ζ jt)δ· · ·Q j∈T1(1 − ζ jt)i and eval-uate it at t = 1.
Consider the following equations g(t) (1 − t)k = Y j∈Tδ (1 − ζjt)δ · · · Y j∈T1 (1 − ζjt) g(t) (1 − t)k = k Y j=1 1 − tdj 1 − t = k Y j=1 1 + t + · · · + tdj−1
2.4. FORMULAS FOR THE OTHER COEFFICIENTS 35 Then, it holds ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) = ∂ ∂t k Y j=1 1 + t + · · · + tdj−1 = = k X i=1 Y j6=1 1 + t + · · · + tdj−1 1 + 2t + · · · + (di− 1)tdi−2 Finally, we have ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 = k X i=1 Y j6=i dj di(di− 1) 2 = = k Y i=1 di· k X i=1 di− 1 2 ! = 1 2· k Y i=1 di· " k X i=1 di− k #
and by substituting this expression in Equation (2.17), we are done.
We are now interested in giving a similar formula for the (k − 3)th coef-ficient of PW
R .
Proposition 2.25Let (R, W ) and PW
R as above. If δ ≤ k − 3, we denote
the (k − 3)th coefficient of Pi, for all i = 0, . . . , d − 1, by ck−3. Then, it holds
ck−3 = 3Pk i=1di 2 −Pk i=1d2i 24(k − 3)!Qk i=1di
Proof. Given i = 0, . . . , d−1, let Pi(n)and S0(n)be as usual. Since δ < k−3
and by remark 2.11, we have that the (k−3)th coefficient of Pi(n)correspond
36 CHAPTER 2. HILBERT QUASI-POLYNOMIAL ck−3 = a0 2(k − 1)!· (k − 1) · k−1 X j=1 j − (k − 1) + +(k − 2) · k−1 X j=1 j − (k − 2) + · · · + 1 · k−1 X j=1 j − (1) + + a1 2(k − 1)!· (k − 2) · k−2 X j=0 j − (k − 2) + +(k − 3) · k−2 X j=0 j − (k − 3) + · · · + 1 · k−2 X j=0 j − (1) + + ... + + ak 2(k − 1)!· (−1) · −1 X j=−k+1 j − (−1) + (−2) · −1 X j=−k+1 j − (−2) + · · · + (−k + 1) · −1 X j=−k+1 j − (−k + 1) = a0 2(k − 1)!· k(k − 1) 2 2 − k−1 X j=1 j2 + + a1 2(k − 1)!· (k − 1)(k − 2) 2 2 − k−2 X j=0 j2 + + ... + + ak 2(k − 1)!· (k − 1)(−k) 2 2 − −1 X j=−k+1 j2 = k X i=0 ai 2(k − 1)! (k − 1)(k − 2i) 2 2 − k−i−1 X j=−i+1 j2
2.4. FORMULAS FOR THE OTHER COEFFICIENTS 37 We observe that for all i = 0, . . . , k it holds
ai 2(k − 1)! (k − 1)(k − 2i) 2 2 − k−i−1 X j=−i+1 j2 = ai 24(k − 1)! ·3k
4+ (−12i − 10)k3+ (12i2+ 36i + 9)k2
+ (−36i2− 24i − 2)k + 24i2]
thus we have ck−3 = 1 24(k − 1)!· " (3k4− 10k3+ 9k2− 2k) · k X i=0 ai+ + (−12k3+ 36k2− 24k) · k X i=0 iai+ + (12k2− 36k + 24) · k X i=0 i2ai # (2.18) We recall that k X i=1 ai = Q0(1) and k X i=1 iai = Q00(1)
and we observe that
k
X
i=1
i2ai = Q000(1) + Q0(1)
By substituting these expressions in Equation 2.19, we obtain ck−3 = 1 24(k − 1)!· [ (3k 4− 10k3+ 9k2− 2k) · Q 0(1)+ +(−12k3+ 36k2− 24k) · Q00(1)+ +(12k2− 36k + 24) · (Q000(1) + Q00(1)) = 1 24(k − 1)!· [ (3k 4− 10k3+ 9k2− 2k) · Q 0(1)+ +(−12k3+ 48k2− 60k + 24) · Q00(1)+ +(12k2− 36k + 24) · Q000(1) ] (2.19) We have already seen that
Q0(1) = 1 Qk i=1di and Q0 0(1) = − Pk i=1di− k 2Qk i=1di
38 CHAPTER 2. HILBERT QUASI-POLYNOMIAL Since Q0(t) = A1(1−t)k−1+A2(1−t)k−2+· · ·+Ak−1(1−t)+Ak, after having
calculated Q00
0(t) and having evaluated it at t = 1, we get Q000(1) = 2Ak−2.
Therefore, it remains to calculate Ak−2. We’ll show in Lemma 2.26 that
Ak−2= 1 24Qk i=1di 3 k X i=1 di− k !2 − k X i=1 (di− 1)(di− 5)
For now let us suppose it holds, and we get Q000(1) = 1 12Qk i=1di 3 k X i=1 di− k !2 − k X i=1 (di− 1)(di− 5)
By substituting the expressions of Q0(1), Q00(1) and Q000(1) in Equation
(2.19), we have ck−3 = 1 24(k − 1)!Qk i=1di · { (3k4− 10k3+ 9k2− 2k)+ +(6k3− 24k2+ 30k − 12) · ( k X i=1 di− k)+ + (k2− 3k + 2) · 3 k X i=1 di− k !2 − k X i=1 (di− 1)(di− 5) = 1 24(k − 3)!Qk i=1di · " k(3k − 1) + 6(k − 1) · k X i=1 di− k ! + + 3 k X i=1 di− k !2 − k X i=1 (di− 1)(di− 5)
After having made calculations, we get ck−3 = 3 ·Pk i=1di 2 −Pk i=1d2i 24 · (k − 3)! ·Qk i=1di
and then we are done.
Lemma 2.26The constant Ak−2 which appears in Equation (2.6) of the expression of Q0(t)is such that
Ak−2 = 1 24 ·Qk i=1di 3 k X i=1 di− k !2 − k X i=1 (di− 1)(di− 5)
2.4. FORMULAS FOR THE OTHER COEFFICIENTS 39 Proof. In order to obtain an expression for Ak−2, we’ll act in a similar way
to which it is done in the calculus of Ak−1. So we are going to evaluate at
t = 1 the second derivative of the Equation (2.7). But before doing it, we observe that the evaluation at t = 1 of the second derivative of the equation gives 0 except for the terms in the first summation for k − 2 ≤ t ≤ k. For this reason, we’ll consider only these addends and we’ll calculate only their derivatives. 0 = ∂ 2 ∂t2 k X t=k−2 (1 − t)k−t· Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 0 = ∂ ∂t (−2Ak−2(1 − t) − Ak−1) · Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt)+ + k X t=k−2 (1 − t)k−t· ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 0 = 2Ak−2 Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt)+ + (−4Ak−2− 2Ak−1) · ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) + + k X t=k−2 (1 − t)k−t· ∂ 2 ∂t2 Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 0 = 2Ak−2· Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 + − 2Ak−1· ∂ ∂t Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 + + Ak· ∂2 ∂t2 Y j∈Tδ (1 − ζjt)δ· · · Y j∈T1 (1 − ζjt) t =1 (2.20)
