Minimal set of binomial generators for certain Veronese 3-fold projections.
Liena Colarte
Department de matemΓ tiques i InformΓ tica, Universitat de Barcelona, Gran Via de les Corts Catalanes 585, 08007 Barcelona,
Spain
[email protected]
Β andΒ
Rosa M. MirΓ³-Roig
Department de matemΓ tiques i InformΓ tica, Universitat de Barcelona, Gran Via de les Corts Catalanes 585, 08007 Barcelona,
Spain
[email protected]
Abstract.
The goal of this paper is to explicitly describe a minimal binomial generating set of a class of lattice ideals, namely the ideal of certain Veronese 3-fold projections. More precisely, for any integer dβ₯4 and any d-th root e of 1 we denote by Xdβ the toric variety defined as the image of the morphism ΟTdββ:P3βΆPΞΌ(Tdβ)β1 where Tdβ are all monomials of degree d in k[x,y,z,t] invariant under the action of the diagonal matrix M(1,e,e2,e3). In this work, we describe a Z-basis of the lattice LΞ·β associated to I(Xdβ) as well as a minimal binomial set of generators of the lattice ideal I(Xdβ)=I+β(Ξ·).
Acknowledgments: The authors are partially supported
by MTM2016β78623-P.
Key words and phrases. Monomial ideals, binomial ideals, lattice ideals, GT-systems,
toric varieties.
2010 Mathematic Subject Classification. 13E10, 14M25, 14N05, 14N15, 53A20.
Contents
- 1 Introduction
- 2 Preliminaries
- 3 GT-systems and GT-varieties
- 4 The lattice of a GT-variety
- 5 A minimal set of generators for GT-lattice ideals
- 6 Final remarks and open problems
1. Introduction
A binomial ideal Iβk[x0β,β¦,xnβ] with k a field
is an ideal generated by polynomials with at most two summands, say axΞ±+bxΞ², where a,bβk and Ξ±,Ξ²βZ+n+1β. Binomial ideals are a large class of ideals which have been amply studied in Combinatoric, Commutative Algebra as well as in Algebraic Geometry. In [11], it was stated that prime binomial ideals are precisely the defining ideals of toric varieties and hence they are lattice ideals, i.e. given a prime binomial ideal Iβk[x0β,β¦,xnβ] there is a lattice LβZn+1 such that I=ILβ:={xuβxvβ£u,vβZn+1Β andΒ uβvβL}.
Ever since, to compute explicitly a minimal set of generators for lattice ideals has been a challenging problem. It is worthwhile to point out that for a given generating set D of the lattice L the ideal I(D)=(xΞ±+βxΞ±βββ£Ξ±+β,Ξ±βββZ+n+1β,Ξ±+ββΞ±βββD)βILβ and the equality does not hold in general.
In [11], Eisenbud and Sturmfels proved that ILβ is a prime ideal if and only if the lattice L is saturated. For prime binomial ideals, a set of generators D of L completely characterizes a set of generators of ILβ. Indeed, a generating set D of L is called a Markov basis if for any lattice point Ξ±+ββΞ±βββL there exits a finite sequence {a1β,β¦,atβ}βZ+n+1β satisfying a1β=Ξ±+β,atβ=Ξ±ββ and aiββai+1ββD for all 1β€iβ€t. In [9], Diaconis and Sturmfels showed that given a set of generators D of L then I(D)=ILβ if and only if D is a Markov basis. We cite [5], [6], [7], [9] and [14] for a detailed exposition of Markov bases of lattice ideals and related problems.
In this paper, we focus our attention in computing a minimal binomial set of generators of a large family of binomial ideals I(Xdβ). They are the ideals associated to suitable projections of Veronese 3-folds.
A Veronese 3-fold V is a projective variety given parametrically by the set M3,dβ of all monomials of degree d in k[x0β,x1β,x2β,x3β] and by a projection of V we understand a projective 3-fold given parametrically by a subset of M3,dβ. In [10], GrΓΆbner proved that V is arithmetically Cohen-Macaulay (aCM, for short)
and its ideal I(V) is generated by quadrics. This is not longer true for all projections of V and it is a longstanding open problem to find a minimal set of generators of any projection of V as well as determine whether a projection of V is aCM. In this paper, we will consider as a subset of M3,dβ the set Tdβ of all monomials of degree d invariant under the action of the diagonal matrix M(1,e,e2,e3) where e is a primitive root of 1 of order d.
Our interest in these ideals Tdβ relies on the following three facts: (1) For all dβ₯4 Tdβ fails the Weak Lefschetz property (WLP) in degree dβ1; (2) The associated morphism ΟTdββ:P3βΆPΞΌ(Tdβ)β1 is a Galois cover of degree d with cyclic Galois group Z/dZ and the image Xdβ of ΟTdββ is a 3-dimensional rational projective variety smooth outside the image of the 4 fundamental points. We call it a GT-threefold; and (3) the 3-fold
Ydβ=Im(Ο)β
where
Ο:Pnβ’P(d3+dβ)βΞΌ(Tdβ)β1 is the rational map associated to (Iβ1)dβ,
satisfies at least one Laplace equation of order
dβ1.
Our goal is to prove that the homogeneous ideal I(Xdβ) of the GT-threefold Xdβ is the homogeneous prime binomial ideal associated to a saturated partial character Ξ· of ZΞΌ(Tdβ) with associated lattice LΞ·β. Afterwards we explicitly compute a minimal binomial set of generators of I(Xdβ). The lattice points associated to these set of generators form a Markov basis of LΞ·β. Our main result states that I(Xdβ) is generated by quadrics if d is even and by quadrics and cubics if d is odd.
Next we outline the structure of this note. In Section 2, we fix the notation we use in the rest of this paper, we relate artinian ideals failing the Weak Lefschetz Property to projective varieties satisfying at least one Laplace equation and we recall the notion of Togliatti systems and GT-systems introduced in [19] and [17]. In Section 3, we give an explicit description of all monomials Tdβ, dβ₯4, invariant under the action of the diagonal matrix M(1,e,e2,e3) and we prove that Tdβ is a GT-system (Proposition 3.3).
The main body of this work is developed in Sections 4 and 5. We denote by Xdβ the GT-threefold associated to the GT-system Tdβ and we first show that Xdβ is an irreducible toric variety whose associated ideal I(Xdβ) is a lattice ideal. In section 4, we consider the
ideal Idβ generated by all binomials of degree 2 vanishing in Xdβ. We associate to Idβ a lattice LΞ·β and a partial character Ξ· of ZΞΌ(Tdβ). We demonstrate that LΞ·β is a saturated lattice of rank ΞΌ(Idβ)β4 (Theorem 4.3) and we show that I(Xdβ) is the lattice ideal I+β(Ξ·) of LΞ·β (Corollary 4.4). In Section 4, we also describe a Z-basis of the lattice LΞ·β (Corollary 4.16) and we explore the relation between Idβ and I+β(Ξ·).
We devote Section 5 to explicitly determine a minimal set of generators of the lattice ideals
I(Xdβ). Our main result states that Idβ=I(Xdβ) if d is even and I(Xdβ)=Idβ+J if d is odd where J is an ideal generated by certain set of cubics of I(Xdβ) that we properly specify (Theorem 5.6). All techniques and results we develop to study the lattice ideal I(Xdβ) are inspired by the ones of Markov basis explained in [9], [14] and [7]. The set of lattice points of generators of Idβ if d even and Idβ and J if d odd forms a Markov basis of LΞ·β. In Section 6, we observe that all GT-varieties are aCM and we concern about computing a minimal free resolution of Xdβ.
Acknowledgement. The authors would like to thank E. Mezzetti and M. Salat for useful discussions on Galois-Togliatti systems.
2. Preliminaries
Throughout this paper we consider the homogeneous polynomial ring R=k[x0β,β―,xnβ] where k is an algebraically closed field of characteristic zero.
Let IβR be a homogeneous artinian ideal. We say that I has the Weak Lefschetz Property (WLP)
if there is a linear form Lβ(R/I)1β such that, for all
integers j, the multiplication map
[TABLE]
has maximal rank, i.e.Β it is injective or surjective. Though many homogeneous artinian ideals are expected to have the
WLP, establishing this property is often rather difficult.
Recently the failure of the WLP has been connected to a large number of problems which seem to be unrelated at first glance.
For example, in [19], Mezzetti, MirΓ³-Roig and Ottaviani proved
that the failure of the WLP is related to the existence of varieties satisfying at least one Laplace
equation of order greater than 2. More precisely, they proved:
Theorem 2.1**.**
Let IβR be an artinian
ideal
generated
by r forms F1β,β¦,Frβ of degree d and let Iβ1 be its Macaulay inverse system.
If
rβ€(nβ1n+dβ1β), then
the following conditions are equivalent:
I* fails the WLP in degree dβ1;*
F1β,β¦,Frβ* become
k-linearly dependent on a general hyperplane H of Pn;*
the n-dimensional variety
X=Im(Ο)β
where
Ο:Pnβ’P(dn+dβ)βrβ1 is the rational map associated to (Iβ1)dβ,
satisfies at least one Laplace equation of order
dβ1.
Proof.
See [19, Theorem 3.2].
β
Motivated by the above results, Mezzetti, MirΓ³-Roig and Ottaviani introduced the following definitions (see [19] and [17]):
Definition 2.2**.**
Let IβR be an artinian ideal generated by r forms of degree d, and rβ€(nβ1n+dβ1β). We will say:
I is a Togliatti system if it fails the WLP in degree dβ1.
I is a monomial Togliatti system if, in addition, I can be generated
by monomials.
I is a smooth Togliatti system if, in addition, the rational variety X is smooth.
A monomial Togliatti system I is minimal if there is no proper subset of the set of generators defining a monomial Togliatti system.
The names are in honor of Togliatti who classified all
rational surfaces parameterized by cubics and satisfying at least one Laplace equation of
order 2 and he proved that for
n=2 the only smooth
Togliatti system of cubics is
[TABLE]
(see [3], [22] and [23]). The systematic study of Togliatti systems was initiated in [19]
and for recent results the reader can see [20], [17], [1], [21] and [18].
Precisely in the latter reference the authors introduced the notion of GT-system which we recall now.
Definition 2.3**.**
A GT-system is an artinian ideal IβR generated by r forms F1β,β¦,Frβ of degree d
such that:
- i)
I* is a Togliatti system.*
2. ii)
The regular map ΟIβ:PnβPrβ1 defined by (F1β,β¦,Frβ) is a Galois covering of degree d with cyclic Galois group Z/dZ.
Any representation of the cyclic group Z/dZ as subgroup of GL(n+1,k) can be diagonalized.
In particular it is represented by a diagonal matrix of the form
[TABLE]
where e is a primitive dth root of 1 and Ξ±0β,Ξ±1β,β¦,Ξ±nβ are integers with
[TABLE]
It follows (see [8, Proposition 4.6]) that the above definition is equivalent to the next one:
Definition 2.4**.**
Fix integers 3β€dβZ, 2β€nβZ, with nβ€d, and 0β€Ξ±0ββ€Ξ±1ββ€β―β€Ξ±nββ€d, e a primitive d-th root of 1 and MΞ±0β,Ξ±1β,β―,Ξ±nββ a representation of Z/dZ in GL(n+1,k). A GT-system will be an ideal
[TABLE]
generated by all forms of degree d invariant under the action of MΞ±0β,Ξ±1β,β¦,Ξ±nββ provided the number of generators ΞΌ(IΞ±0β,β¦,Ξ±nβdβ)β€(nβ1n+dβ1β).
Finally, note that the ideal IΞ±0β,β¦,Ξ±nβdβ is always monomial,
i.e. a GT-system is a monomial Togliatti system.
3. GT-systems and GT-varieties
Through this section we fix an integer dβ₯4, a dth-root of unity e and we write d=2k+Ξ΅=3kβ²+Ο with Ξ΅β{0,1} and Οβ{0,1,2}. We denote TdββR=k[x,y,z,t] the ideal generated by the ΞΌ(Tdβ) monomials of degree d invariant under the action of the diagonal matrix M(1,e,e2,e3).
In this section, we will describe the ideal Tdβ and we will prove that Tdβ is a GT-system for all dβ₯4. We also define the GT-varieties Xdβ and their apolar varieties Ydβ. The homogeneous ideal I(Xdβ) of a GT-variety Xdβ is a lattice ideal. A basis of the lattice and a system of generators of the lattice ideal will be effectively computed in next sections.
A monomial xΞ±yΞ²zΞ΄tΞ³βR of degree d belongs to Tdβ if it is invariant under the action of
M(1,e,e2,e3) or, equivalently if Ξ±,Ξ²,Ξ΄,Ξ³ satisfy:
[TABLE]
The solutions of (β) in terms of Ξ³ and r are the following:
[TABLE]
Given dβ₯4, we define
[TABLE]
All monomials xΞ±yΞ²zΞ΄tΞ³βTdβ of degree d are uniquely determined by a triple (r,Ξ³,Ξ΄)βWdβ. In particular,
ΞΌ(Tdβ)=#Wdβ.
Remark 3.1**.**
Notice that xd,yd,zd and td are invariant under the action of M(1,e,e2,e3). So, the ideal Tdβ is artinian.
In next example,
we explicitly exhibit Tdβ for d=4,5,6,7,8 and 9. For these values of d we cover all possibilities of Ξ΅ and Ο.
Example 3.2**.**
\begin{array}[]{rcl}T_{4}&=&(x^{4},y^{4},xy^{2}z,x^{2}z^{2},x^{2}yt,z^{4},yz^{2}t,y^{2}t^{2},xzt^{2},t^{4}),\;\;\mu_{I_{4}}=10.\end{array}
\begin{array}[]{rcl}T_{5}&=&(x^{5},y^{5},xy^{3}z,x^{2}yz^{2},tx^{2}y^{2},tx^{3}z,z^{5},tyz^{3},t^{2}y^{2}z,t^{2}xz^{2},t^{3}xy,t^{5}),\;\;\mu_{I_{5}}=12.\end{array}
\begin{array}[]{rcl}T_{6}&=&(x^{6},y^{6},xy^{4}z,x^{2}y^{2}z^{2},x^{3}z^{3},tx^{2}y^{3},tx^{3}yz,t^{2}x^{4},z^{6},tyz^{4},t^{2}y^{2}z^{2},t^{2}xz^{3},t^{3}y^{3},t^{3}xyz,\\
&&t^{4}x^{2},t^{6}),\;\;\mu_{I_{6}}=16.\end{array}
\begin{array}[]{rcl}T_{7}&=&(x^{7},y^{7},xy^{5}z,x^{2}y^{3}z^{2},x^{3}yz^{3},tx^{2}y^{4},tx^{3}y^{2}z,tx^{4}z^{2},t^{2}x^{4}y,z^{7},tyz^{5},t^{2}y^{2}z^{3},t^{2}xz^{4},t^{3}y^{3}z,\\
&&t^{3}xyz^{2},t^{4}xy^{2},t^{4}x^{2}z,t^{7}),\;\;\mu_{I_{7}}=18.\end{array}
\begin{array}[]{rcl}T_{8}&=&(x^{8},y^{8},xy^{6}z,x^{2}y^{4}z^{2},x^{3}y^{2}z^{3},x^{4}z^{4},tx^{2}y^{5},tx^{3}y^{3}z,tx^{4}yz^{2},t^{2}x^{4}y^{2},t^{2}x^{5}z,z^{8},tyz^{6},t^{2}y^{2}\!z^{4}\!,\\
&&t^{2}xz^{5},t^{3}y^{3}z^{2},t^{3}xyz^{3},t^{4}y^{4},t^{4}xy^{2}z,t^{4}x^{2}z^{2},t^{5}x^{2}y,t^{8}),\;\;\mu_{I_{8}}=22.\end{array}
\begin{array}[]{rcl}T_{9}&=&(x^{9},y^{9},xy^{7}z,x^{2}y^{5}z^{2},x^{3}y^{3}z^{3},x^{4}yz^{4},tx^{2}y^{6},tx^{3}y^{4}z,tx^{4}y^{2}z^{2},tx^{5}z^{3},t^{2}x^{4}y^{3},t^{2}x^{5}yz,t^{3}x^{6},\\
&&z^{9},tyz^{7},t^{2}y^{2}z^{5},t^{2}xz^{6},t^{3}y^{3}z^{3},t^{3}xyz^{4},t^{4}y^{4}z,t^{4}xy^{2}z^{2},t^{4}x^{2}z^{3},t^{5}xy^{3},t^{5}x^{2}yz,t^{6}x^{3},t^{9}),\\
&&\mu_{I_{9}}=26.\end{array}
Our interest in the study of these monomial ideals relies in the following fact:
Proposition 3.3**.**
For any dβ₯4, Tdβ is a GTβsystem. In particular, Tdβ fails the WLP in degree dβ1.
Proof.
By Definition 2.4, we only have to check that ΞΌ(Tdβ)β€(22+dβ). From the definition of Tdβ, it follows that
[TABLE]
We sum separately for r=1 and
r=2; we have
[TABLE]
We only have to focus on the sum of the series of the type βΞ³=1Nββ23Ξ³βΞ΅ββ with
Ξ΅β{0,1}. We can rewrite the series as follows: if N=2j,
βΞ³=1Nββ23Ξ³βΞ΅ββ=βi=1jβ3i+βi=1jβ(3iβ1βΞ΅)=j(3j+2βΞ΅). Otherwise N=2j+1,
βΞ³=1Nββ23Ξ³βΞ΅ββ=βi=1jβ3j+βi=1j+1β3jβ1βΞ΅=(j+1)(3j+2βΞ΅). In any case,
[TABLE]
From this, we conclude
[TABLE]
Substituting d=3kβ²+Ο by k=23kβ²+ΟβΞ΅β we verify that ΞΌ(Tdβ)β€2+(kβ²+1)(23kβ²+Οβ+1)+(3kβ²+Ο+1)(2kβ²+2)+23kβ²+Οβ(23kβ²+Οβ+1)β(3kβ²+Ο)(23kβ²+Οβ+1)β2kβ²β(23(kβ²β1)β+1)βkβ²(3kβ²+2)=41β(20+6(kβ²)2+8ΟβΟ2+kβ²(29+4Ο)). It holds that 41β(20+6(kβ²)2+8ΟβΟ2+kβ²(29+4Ο))<21β(d+2)(d+1)β1/4(16β12(kβ²)2+kβ²(11β8Ο)+2Οβ3Ο2)β€0, which holds
for all dβ₯4.
β
We finish this section studying the geometric properties
of the rational 3-fold associated to the GT-system Tdβ.
The morphism ΟTdββ:P3βΆPΞΌ(Tdβ)β1 associated to the GT-system Tdβ is a Galois cover of degree d with cyclic Galois group Z/dZ represented by M(1,e,e2,e3).
In particular, a general fibre of ΟTdββ consists of d points, and hence the image of ΟTdββ is a 3-dimensional rational projective variety.
Definition 3.4**.**
We call GT-variety and we denote it by Xdβ the rational 3-fold defined as the image of ΟTdββ.
The morphism
ΟTdββ:P3βΆPΞΌ(Tdβ)β1 is unramified outside the four fundamental points of
P3: E0β=[1,0,0,0], E1β=[0,1,0,0], E2β=[0,0,1,0] and E4β=[0,0,0,1]. They are sent by
ΟTdββ to the singular points of Xdβ, Piβ:=Ο(Eiβ),
i=0,1,2,3, that are cyclic quotient singularities: P0β is of type d1β(1,2,3), P1β is of type 21β(1,dβ1,dβ2), P2β is of type d1β(dβ2,dβ1,1) and
P3β is of type d1β(dβ3,dβ2,dβ1).
Remark 3.5**.**
(1)
It is worthwhile to point out that the rational 3-fold Xdβ is also a Galois covering of P3 with Galois
group Z/dZ. The covering map XdββP3 composed with ΟTdββ
is P3βP3, [x,y,z,t]β[xd,yd,zd,td].
(2) Let Tdβ1β be the Macaulay inverse system of Tdβ and denote by Ydβ the rational 3-fold defined as the closure of the image of the rational map
ΟTdβ1ββ:P3β’P(d3+dβ)βΞΌ(Tdβ)β1. By Theorem 2.1, Ydβ satisfies a Laplace equation of order dβ1.
Our main goal will be to prove that the homogeneous ideal I(Xdβ) of a GT-variety XdββPΞΌ(Tdβ)β1
is generated by quadrics if d is even and by quadrics and cubics if d is odd (see Corollary 5.7)
4. The lattice of a GT-variety
As in the previous section, we fix dβ₯4 and we write d=2k+Ξ΅=3kβ²+Ο, with Ξ΅β{0,1} and Οβ{0,1,2}.
We want to determine the homogeneous ideal I(Xdβ) of the GT-threefold XdββPΞΌ(Tdβ)β1 defined by the GT-system Tdβ. Since Xdβ is an irreducible toric variety, I(Xdβ) is a binomial ideal of codimension ΞΌ(Tdβ)β4 associated to a lattice LΞ·β. As we already pointed out our main goal is to prove that I(Xdβ) is generated by quadrics if d is even and by quadrics and cubics if d is odd (see Corollary 5.7) but first we will explicitly describe a Z-basis of the lattice LΞ·β associated to I(Xdβ) (see Theorem 4.3).
The ideal Tdβ is generated by the set
{xΞ΄+2Ξ³+(1βr)dyrdβ2Ξ΄β3Ξ³zΞ΄tΞ³β£(r,Ξ³,Ξ΄)βWdβ}βK[x,y,z,t] (see Section 3.1). All these monomials are
uniquely determined by a triple (r,Ξ³,Ξ΄)βWdβ and often we will denote xΞ΄+2Ξ³+(1βr)dyrdβ2Ξ΄β3Ξ³zΞ΄tΞ³ by w(r,Ξ³,Ξ΄)β.
Definition 4.1**.**
We define the binomial ideal Idβ=(w(r1β,Ξ³1β,Ξ΄1β)βw(r2β,Ξ³2β,Ξ΄2β)ββw(r3β,Ξ³3β,Ξ΄3β)βw(r4β,Ξ³4β,Ξ΄4β)ββ£ r1β+r2β=r3β+r4β,Ξ³1β+Ξ³2β=Ξ³3β+Ξ³4β,Ξ΄1β+Ξ΄2β=Ξ΄3β+Ξ΄4β)βk[w(r,Ξ³,Ξ΄)β](r,Ξ³,Ξ΄)βWdββ.
Let us illustrate the above definition with an example.
Example 4.2**.**
We take d=4,(k=2,kβ²=1,Ξ΅=0,Ο=1). We have (Example 3.2):
[TABLE]
and
[TABLE]
Solving the equation (r1β,Ξ³1β,Ξ΄1β)+(r2β,Ξ³2β,Ξ΄2β)=(r3β,Ξ³3β,Ξ΄3β)+(r4β,Ξ³4β,Ξ΄4β) in W4β we obtain twelve generators for I4β:
[TABLE]
By construction it follows that Idβ vanishes on Xdβ, and hence IdββI(Xdβ). Let k[w(r,Ξ³,Ξ΄)Β±β] be the ring of Laurent polynomials over k. To each binomial
[TABLE]
we associated a Laurent binomial
[TABLE]
They generate a Laurent binomial ideal whose associated partial character is the trivial one
Ξ·:LΞ·ββkβ, sending Ξ·(m)=1 for all mβLΞ·β, where LΞ·β=β¨Ξ±β£wΞ±+ββwΞ±βββIdββ©. In turn, the partial character Ξ· induces a lattice ideal I+β(Ξ·)=(wΞ±+ββwΞ±βββk[w(r,Ξ³,Ξ΄)β]β£Ξ±βLΞ·β).
Now we state the main result of this section.
Theorem 4.3**.**
- (1)
The lattice LΞ·β is saturated and rk(LΞ·β)=ΞΌ(Tdβ)β4.
2. (2)
I+β(Ξ·)=(βi=1nβw(riβ,Ξ³iβ,Ξ΄iβ)βββi=1nβw(riβ²β,Ξ³iβ²β,Ξ΄iβ²β)ββk[w(r,Ξ³,Ξ΄)β]β£βi=1nβriβ=βi=1nβriβ²β,βi=1nβΞ³iβ=βi=1nβΞ³iβ²β,βi=1nβΞ΄iβ=βi=1nβΞ΄iβ²β}.
Corollary 4.4**.**
I(Xdβ)=I+β(Ξ·).
Proof.
Theorem 4.3 (1) implies that I+β(Ξ·) is a prime ideal of codimension 4 (see [11, Corollary 2.5 and 2.6]).
From Theorem 4.3 (2) it follows that I+β(Ξ·) vanishes in Xdβ, i.e. I+β(Ξ·)βI(Xdβ). Therefore, I+β(Ξ·) is the homogeneous ideal of an irreducible 3-dimensional variety contained in Xdβ. Since Xdβ is irreducible we conclude that I+β(Ξ·)=I(Xdβ) which proves what we want.
β
We trivially have IdββI+β(Ξ·)=I(Xdβ). In next section we will discuss whether the equality holds. Now we devote the rest of this section to prove Theorem 4.3.
Definition 4.5**.**
Fixed nβ₯2, we define a suitable n-binomial to be a nonzero binomial wΞ±=wΞ±+ββwΞ±ββ=βi=1nβw(riβ,Ξ³iβ,Ξ΄iβ)βββi=1nβw(riβ²β,Ξ³iβ²β,Ξ΄iβ²β)β satisfying βi=1nβriβ=βi=1nβriβ²β, βi=1nβΞ³iβ=βi=1nβΞ³iβ²β and βi=1nβΞ΄iβ=βi=1nβΞ΄iβ²β.
Remark 4.6**.**
Any suitable n-binomial wΞ± vanishes in Xdβ. Therefore, all suitable n-binomials belong to I(Xdβ). Moreover, the generators of
Idβ are suitable 2-binomials.
Definition 4.7**.**
Given a suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ, we note supp+β(wΞ±) (respectively suppββ(wΞ±)) the support of the monomial wΞ±+β (respectively
support of wΞ±ββ). We say that wΞ± is non trivial if supp+β(wΞ±)β©suppββ(wΞ±)=β
. Otherwise, we say that wΞ± is trivial.
Example 4.8**.**
The set of generators for I4β in Example 4.2 are the set of all non-trivial suitable 2-binomials.
Definition 4.9**.**
Let m=βi=1nβw(riβ,Ξ³iβ,Ξ΄iβ)ββk[w(r,Ξ³,Ξ΄)β] be a monomial of degree n. We say that m admits a suitable n-binomial if there exists a monomial mβ²=βi=1nβw(riβ²β,Ξ³iβ²β,Ξ΄iβ²β)ββk[w(r,Ξ³,Ξ΄)β] of degree n such that mβmβ² is a non trivial suitable n-binomial.
Let us order the elements (r,Ξ³,Ξ΄)βWdβ lexicographically.
Definition 4.10**.**
We say that w(r,Ξ³,Ξ΄)ββk[w(r,Ξ³,Ξ΄)β] admits a special n-binomial if there exists a non trivial suitable n-binomial mβmβ²βI+β(Ξ·) such that (r,Ξ³,Ξ΄)=min{supp(mβmβ²)}.
Example 4.11**.**
The element w(0,0,0)ββk[w(r,Ξ³,Ξ΄)β] admits a special 2-binomial. Indeed, w(0,0,0)βw(2,2kβ²,0)ββw(1,kβ²,0)βw(1,kβ²,0)β is a non trivial suitable 2-binomial and (0,0,0)=min{(0,0,0), (2,2kβ²,0),(1,kβ²,0)}. While clearly the element w(3,d,0)β does not admit a special n-binomial for any nβ₯2.
Example 4.12**.**
For d=4, the set of elements admitting a special 2-binomial is W4ββ{(1,1,0),(2,1,2),(2,2,0), (2,2,1),(3,4,0)} while the element (1,1,0) admits a special 3-binomial: w(1,1,0)βw(2,1,2)βw(3,4,0)ββw(2,2,0)βw(2,2,1)2β.
Lemma 4.13**.**
Each monomial m=w(1,Ξ³,Ξ΄)βw(3,d,0)ββk[w(r,Ξ³,Ξ΄)β] admits a special 2-binomial except: (Ξ³,Ξ΄)=(kβ²,β2Οββ) if Οξ =0, and Ξ³=Ξ΄=0 if Ξ΅=1.
Proof.
Fix (1,Ξ³,Ξ΄)βWdβ. If there exists such monomial mβ², it has to be of the form
w(2,Ξ³1β,Ξ΄1β)βw(2,Ξ³2β,Ξ΄2β)β with
0β€Ξ³iββ€2kβ²+β32Οββ, max{0,dβ2Ξ³iβ}β€Ξ΄iββ€β22dβ3Ξ³iβββ, i=1,2, Ξ³+d=Ξ³1β+Ξ³2β and Ξ΄=Ξ΄1β+Ξ΄2β. From this follows that when Ο=1 and Ξ³=kβ², there are no Ξ³1β and Ξ³2β summing Ξ³+d=4kβ²+1. While for Ο=2, we must have Ξ³1β=Ξ³2β=2kβ²+1. But then Ξ΄1β=Ξ΄2β=0, which cannot sum Ξ΄=1.
For the rest of Ξ³βs, we set Ξ³1β:=β2d+Ξ³ββ and Ξ³2β:=β2d+Ξ³ββ.
Observe that we always have kβ€Ξ³1β,Ξ³2ββ€2kβ²+β2Οββ. From the properties of the floor and ceiling functions we have
[TABLE]
where the equality holds when Ξ³1β and Ξ³2β are not both odd. If the equality holds we can find values Ξ΄1β and Ξ΄2β such that Ξ΄1β+Ξ΄2β=Ξ΄, as long as
Ξ΄β₯max{0,dβ2Ξ³1β}+max{0,dβ2Ξ³2β}. The last condition always happens except for Ξ³=Ξ΄=0 when Ξ΅=1.
Finally, if Ξ³1β and Ξ³2β are odd (and, hence,
Ξ³β₯2), the result follows taking mβ²=w(2,Ξ³1β+1,β22dβ3(Ξ³1β+1)ββ)βw(2,Ξ³2ββ1,β22dβ3(Ξ³2ββ1)ββ)β.
β
Proposition 4.14**.**
All w(1,Ξ³,Ξ΄)ββk[w(r,Ξ³,Ξ΄)β] admit a special 2-binomial or 3-binomial.
Proof.
It is enough to treat the 3 exemptions of Lemma 4.13. For Ξ΅=1 and (1,Ξ³,Ξ΄)=(1,0,0) it is enough to observe that
w(1,0,0)βw(2,2kβ²,0)ββw(1,1,0)βw(2,2kβ²β1,0)β if Ο=0, w(1,0,0)βw(2,2kβ²,1)ββw(1,0,1)βw(2,2kβ²,0)β if Ο=1 and
w(1,0,0)βw(2,2kβ²+1,0)ββw(1,1,0)βw(2,2kβ²,0)β if Ο=2 are special 2-binomials.
For (1,Ξ³,Ξ΄)=(1,kβ²,β2Οββ) and Οξ =0, the monomial w(1,kβ²,β2Οββ)β does not admit a special 2-binomial. However, w(1,kβ²,0)βw(2,2kβ²β1,2)βw(3,d,0)ββw(2,2kβ²,0)βw(2,2kβ²,1)2β for Ο=1 and w(1,kβ²,1)βw(2,2kβ²,1)β w(3,d,0)ββw(2,2kβ²,2)βw(2,2kβ²+1,0)2β for Ο=2 are special 3-binomials.
β
Proposition 4.15**.**
All w(2,Ξ³,Ξ΄)ββk[w(r,Ξ³,Ξ΄)β] admit a special 2-binomial or 3-binomial except
{w(2,2kβ²β1,0)β,w(2,2kβ²β1,1,)β,w(2,2kβ²,0)β} if Ο=0, {w(2,2kβ²β1,2)β,w(2,2kβ²,0)β,w(2,2kβ²,1)β} if Ο=1, and
{w(2,2kβ²,1)β,w(2,2kβ²,2)β,w(2,2kβ²+1,0)β} if Ο=2.
Proof.
For any (2,Ξ³,Ξ΄)βWdβ different from the excluded cases we consider the monomial m=w(2,Ξ³,Ξ΄)βw(2,2kβ²+β2Οββ,β2Οββββ2Οββ)β. For convenience we note
Ξ³β²=2kβ²+β2Οββ and Ξ΄β²=β2Οββββ2Οββ. Set Ξ³1β:=Ξ³+1 and Ξ³2β:=Ξ³β²β1. Unless Ξ³ and Ξ³β² are even, and Ξ΄=(2dβ3Ξ³)/2 (hence Οξ =2), there exists
Ξ΄iβ with max{0,dβ2Ξ³iβ}β€Ξ΄iββ€β22dβ3Ξ³iβββ such that
Ξ΄1β+Ξ΄2β=Ξ΄+Ξ΄β².
If Ξ³ and Ξ³β² are even, Ξ΄=(2dβ3Ξ³)/2 and Ξ³<2kβ²β2 we take Ξ³1β:=Ξ³+2 and Ξ³2β:=2kβ²β2. Then, there exists
Ξ΄iβ with max{0,dβ2Ξ³iβ}β€Ξ΄iββ€β22dβ3Ξ³iβββ such that
Ξ΄1β+Ξ΄2β=Ξ΄+Ξ΄β².
If Ξ³=2kβ²β2 and Ο=1, w(2,2kβ²β2,4)βw(2,2kβ²,0)ββw(2,2kβ²β1,2)2β is a special 2-binomial when Ο=1.
Finally, if Ο=0, Ξ³=2kβ²β2 and Ξ΄=3, the element (2,2kβ²β2,3) does not admit a special 2-binomial but it admits a special 3-binomial:
w(2,2kβ²β2,3)βw(2,2kβ²β1,0)βw(2,2kβ²,0)ββw(2,2kβ²β1,1)3β.
β
From now on we set:
Wdβ²β=Wdββ{(2,2kβ²β1,0),(2,2kβ²β1,1),(2,2kβ²,0),(3,d,0)} if Ο=0,
Wdβ²β=Wdββ{(2,2kβ²β1,2),(2,2kβ²,0),(2,2kβ²,1),(3,d,0)} Ο=1, and
Wdβ²β=Wdββ{(2,2kβ²,0),(2,2kβ²,1),(2,2kβ²,2),(3,d,0)} Ο=2.
Up to now we have seen that for any (r,Ξ³,Ξ΄)βWdβ²β the variable w(r,Ξ³,Ξ΄)β admits a special 2-binomial or 3-binomial.
For each (r,Ξ³,Ξ΄)βWdβ²β set D(r,Ξ³,Ξ΄)β to be one of its special binomials and note Ξ±(r,Ξ³,Ξ΄)β its lattice point. We call {D(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ a system of special binomials and {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ its associated system of lattice points. The matrix associated to any system of lattice points is upper triangular. So we have the following result:
Corollary 4.16**.**
For any system of special binomials {D(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ its associated system of lattice points {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ is a Z-basis of ZΞΌ(Tdβ)β4.
Example 4.17**.**
For d=4 we can chose as a system of special binomials
[TABLE]
The matrix associated to its system of lattice points is
[TABLE]
So, {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βW4β²ββ is Z-basis of Z6.
Next we prove that any system of lattice points is a Z-basis of
the lattice LΞ·β. In the sequel we fix {D(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ and its associated system of lattice points {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ. Rephrasing, we want to demonstrate that LΟβ=β¨Ξ±(r,Ξ³,Ξ΄)ββ©(r,Ξ³,Ξ΄)βWdβ²ββ.
The lattice LΞ·β is generated by all suitable 2-binomials. Thus it is enough to express the lattice point of any non-trivial suitable 2-binomial as a linear combination of {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²ββ. So we fix
a non-trivial suitable 2-binomial
[TABLE]
with associated lattice point
[TABLE]
Set Ξ±1β:=Ξ±0βββWdβ²β,Ξ±0β+βΞ±(riβ,Ξ³iβ,Ξ΄iβ)β+βWdβ²β,Ξ±0βββΞ±(riβ,Ξ³iβ,Ξ΄iβ)β, where the summing βWdβ²β,Ξ±0β+β (respectively, βWdβ²β,Ξ±0βββ) means that we only consider those elements (riβ,Ξ³iβ,Ξ΄iβ)βWdβ²ββ©supp(Ξ±0+β) (respectively, Wdβ²ββ©supp(Ξ±0ββ)). Therefore, Ξ±1β is a point of LΞ·β and its associated binomial wΞ±1β is a suitable n-binomial for some nβ₯2.
Furthermore, supp(Ξ±0β)β©supp(Ξ±1β)β©Wdβ²β=β
and all elements in supp(Ξ±1β) are strictly bigger than min{supp(Ξ±0β)}.
If there exists a lattice point (r,Ξ³,Ξ΄)βWdβ²ββ©supp(Ξ±1β), then we apply the same strategy to Ξ±1β and so on. Before continuing let us see how the procedure works by an example.
Example 4.18**.**
According to Example 4.2 for d=4 we have 12 non-trivial suitable 2-binomials. Five of them are part of the system of special binomials that we fix in Example 4.17. Let us check that the seven remaining cases can be written as a linear combination of the system of special binomials fixed in Example 4.17. The first step of the above induction process gives us:
[TABLE]
Since D(2,0,4)β=w(2,0,4)βw(2,2,0)ββw(2,1,2)2β (see Example 4.17), next step reduces Ξ±1β to [math] in all cases.
In general, this procedure defines inductively a sequence of lattice points
{Ξ±1β,β¦,Ξ±sβ,β¦} βLΞ·β, such that at any step s of the induction process supp(Ξ±sβ1β)β©supp(Ξ±sβ)β©Wdβ²β=β
and min{supp(Ξ±sβ)} is strictly smaller than any element in the support of Ξ±sβ1β. So clearly this process stops, indeed Wdβ²β is finite. Once it ends, we obtain a linear combination of {Ξ±(r,Ξ³,Ξ΄)β}(r,Ξ³,Ξ΄)βWdβ²βββͺ{Ξ±0β}, we denote it Ξ±hββLΞ·β for some hβ₯1. To achieve our goal it suffices to check that Ξ±hβ=0. We note the elements of WdββWdβ²β by l1β,l2β,l3β and l4β, ordered in the natural way. It is a matter of fact that wΞ±hβ is a suitable n-binomial and supp(Ξ±hβ)βWdββWdβ²β. Thus there exist non negative integers A1β,β¦,A4β and B1β,β¦,B4β such that Ξ±hβ=βi=14βAiβliβββi=14βBiβliβ.
Lemma 4.19**.**
With the above notation, (3,d,0)β/supp(Ξ±hβ).
Proof.
Since wΞ±hβ is a suitable n-binomial, it holds that 2(A1β+A2β+A3β)+3A4β=2(B1β+B2β+B3β)+3B4β. In other words, the rβs involved in supp(Ξ±h+β) and supp(Ξ±hββ) form a two full partitions of the same length A1β+A2β+A3β+A4β=B1β+B2β+B3β+B4β and weight 2(A1β+A2β+A3β)+3(A4β). So necessarily A4β=B4β which proves what we want.
β
For sake of completeness we specify Ξ±hβ in each case.
Ξ±hβ=(A1ββB1β)(2,2kβ²β1,0)+(A2ββB2β)(2,2kβ²β1,1)+(A3ββB3β)(2,2kβ²,0) when Ο=0;
Ξ±hβ=(A1ββB1β)(2,2kβ²β1,2)+(A2ββB2β)(2,2kβ²,0)+(A3ββB3β)(2,2kβ²,1) if Ο=1; and
Ξ±hβ=(A1ββB1β)(2,2kβ²+1,0)+(A2ββB2β)(2,2kβ²,1),+(A3ββB3β)(2,2kβ²,2) for Ο=2.
Since wΞ±hβ is a suitable n-binomial, a straightforward computation shows that Aiβ=Biβ, i=1,2,3.
β‘
5. A minimal set of generators for GT-lattice ideals
In the previous section we have stated that I(Xdβ) is a lattice ideal and we have given a Z-basis of the associated lattice LΞ·β as well as a
system of generators of I(Xdβ) ( Theorem 4.3 (2)). Precisely, I(Xdβ) is generated by all non trivial suitable n-binomials with nβ₯2. Now we want to determine a minimal set of generators for I(Xdβ). More concretely,
we will prove that the GT-lattice ideal I(Xdβ)
is generated by quadrics if d is even and by quadrics and cubics if d is odd (Corollary 5.7).
As in previous sections dβ₯4 and we write d=2k+Ξ΅=3kβ²+Ο, with Ξ΅β{0,1} and Οβ{0,1,2}.
For each nβ₯2 we denote I+β(Ξ·)nβ the set of all suitable n-binomials and
(I+β(Ξ·)nβ) the ideal of k[w(r,Ξ³,Ξ΄)β] generated by them. Therefore, we have
[TABLE]
Definition 5.1**.**
Let wΞ±=wΞ±+ββwΞ±ββ be a non trivial suitable n-binomial. By an I+β(Ξ·)nβ-sequence from wΞ±+β to wΞ±ββ we mean a finite sequence {wa1β,β¦,watβ} of monomials in k[w(r,Ξ³,Ξ΄)β] satisfying the following two conditions:
wa1β=wΞ±+β,watβ=wΞ±ββ and
For all 1β€j<t, wajββwaj+1β is a trivial suitable n-binomial.
The second condition in the above definition says that for each 1β€j<t, there exists a variable w(rjβ,Ξ³jβ,Ξ΄jβ)ββsupp(wajβ)β©supp(waj+1β). Thus each wajββwaj+1β belongs to (I+β(Ξ·)nβ1β).
Example 5.2**.**
Any trivial suitable n-binomial wΞ±+βwΞ±β gives rise to the I+β(Ξ·)nβ-sequence {wΞ±+β,wΞ±ββ}.
Example 5.3**.**
Consider d=4 and I4β from Example 4.2. The lattice ideal I4β is generated by all suitable 2-binomials. Let us give some examples of I+β(Ξ·)3β-sequence. Set wa1β=w(0,0,0)βw(1,0,2)βw(2,1,2)β. Since w(1,0,2)βw(2,1,2)ββw(1,1,0)βw(2,0,4)β is a suitable 2-binomial, {wa1β,wa2β} with wa2β:=w(0,0,0)βw(1,1,0)βw(2,0,4)β is an I+β(Ξ·)3β-sequence. Now observe that w(0,0,0)βw(2,0,4)ββw(1,0,2)βw(1,0,2)β is also a suitable 2-binomial.
Hence wa2ββwa3β with wa3β:=w(1,1,0)βw(1,0,2)βw(1,0,2)β is trivial and so {wΞ±1β,wa2β,wa3β} is an I+β(Ξ·)3β-sequence from wΞ±1β to wΞ±3β.
As another example of I+β(Ξ·)3β-sequence we have
[TABLE]
and the equality
[TABLE]
shows that the non trivial 3-binomial w(1,0,0)βw(1,0,2)βw(2,2,1)ββw(1,0,1)βw(1,1,0)βw(2,1,2)ββ(I+β(Ξ·)2β).
This last example illustrates very well what happens in general. Indeed, we have:
Proposition 5.4**.**
Fix nβ₯3 and let wΞ±=wΞ±+ββwΞ±ββ be a suitable n-binomial. Then wΞ±β(I+β(Ξ·)nβ1β) if and only if there exists an I+β(Ξ·)nβ-sequence from wΞ±+β to wΞ±ββ.
Proof.
Suppose that wΞ±β(I+β(Ξ·)nβ1β). We note I+β(Ξ·)nβ1β :={q1β,β¦,qNβ} with N the number of all suitable (nβ1)-binomials and qjβ=qju+jβββqjuβjββ. By hypothesis there exist homogeneous linear forms l1β,β¦,lNβ such that wΞ±+β=l1βq1β+β―+lNβqNβ+wΞ±ββ. Now we write ljβ=a(0,0,0)jβw(0,0,0)β+β―+a(3,d,0)jβw(3,d,0)β, where a(r,Ξ³,Ξ΄)jββk for all (r,Ξ³,Ξ΄)βWdβ and j=1,β¦,N. Therefore wΞ±+β=βjNββ(r,Ξ³,Ξ΄)βWdββ(a(r,Ξ³,Ξ΄)jβw(r,Ξ³,Ξ΄)βqju+jβββa(r,Ξ³,Ξ΄)jβw(r,Ξ³,Ξ΄)βqjuβjββ)+wΞ±ββ. Hence, there exists j0β such that a(r0β,Ξ³0β,Ξ΄0β)j0ββ=1 and
wΞ±+β=w(r0β,Ξ³0β,Ξ΄0β)βqj0βu+j0βββ
or a(r0β,Ξ³0β,Ξ΄0β)j0ββ=β1 and
wΞ±+β=w(r0β,Ξ³0β,Ξ΄0β)βqj0βuβj0βββ . Assume a(r0β,Ξ³0β,Ξ΄0β)j0ββ=1 (analogously we deal with the case a(r0β,Ξ³0β,Ξ΄0β)j0ββ=β1). Set
wa2β=w(r0β,Ξ³0β,Ξ΄0β)βqj0βuβj0βββ. We have
[TABLE]
Thus
[TABLE]
We iterate the process, first with wa2β, we construct the I+β(Ξ·)nβ-sequence; and taking into account that the number of summands decreases at each step we can assure that we end with what
we are looking for. We only have to note that the described process stops, since at each step we reduce the number of members of the linear combination, which is finite.
Therefore watβ=wΞ±ββ for some t>2.
β
Let m be the smallest integer mβ₯2 such that
any suitable (m+1)-binomial of I+β(Ξ·)m+1β admits a I+β(Ξ·)(m+1)β-sequence. By (4.4) and Proposition 5.4 we have
[TABLE]
Notation 5.5**.**
For any odd integer dβ₯5, we define
M30β:={w(0,0,0)βw(2,0,d)βw(1,0,Ξ΄)β}Ξ΄=0kβ1ββͺ{w(1,0,0)βw(2,Ξ³,dβ2Ξ³)βw(3,d,0)β}Ξ³=0kβ1β and
M31β=M32β:={w(0,0,0)βw(2,0,d)βw(1,0,Ξ΄)β}Ξ΄=0kβ1ββͺ{w(1,0,0)βw(2,Ξ³,dβ2Ξ³)βw(3,d,0)β}Ξ³=0kβ1ββͺ{w(0,0,0)βw(2,0,d)βw(3,d,0)β, w(0,0,0)βw(1,0,0)βw(3,d,0)β}.
Now we state our main result.
Theorem 5.6**.**
If d is even, for any nβ₯3 and any suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ there exists a I+β(Ξ·)nβ-sequence from wΞ±+β to
wΞ±ββ.
If d is odd, for any nβ₯4 and any suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ there exists a I+β(Ξ·)nβ-sequence from wΞ±+β to
wΞ±ββ.
If d is odd and n=3 then a suitable 3-binomial wΞ±=wΞ±+ββwΞ±ββ admits a I+β(Ξ·)3β-sequence from wΞ±+β to
wΞ±ββ if and only if neither wΞ±+β nor wΞ±ββ belong to M3Οβ.
Corollary 5.7**.**
(1) If dβ₯4 is even, then I+β(Ξ·)=(I+β(Ξ·)2β)=Idβ.
(2) If dβ₯5 is odd, then I+β(Ξ·)=(I+β(Ξ·)2β)+(I+β(Ξ·)3β)=Idβ+(wΞ±βI+β(Ξ·)3ββ£wΞ±+ββM3ΟβorwΞ±βββM3β)Ο.
We devote the rest of this section to prove Theorem 5.6 but first let us illustrate it with a couple of examples.
Example 5.8**.**
Using the software Macaulay2, we check that I(X4β)=T4β (see Example
4.2).
Example 5.9**.**
Fix d=5, the binomial ideal I5β is generated by twenty suitable 2-binomials, all lattice points satisfying the equation (r1β,Ξ³1β,Ξ΄1β)+(r2β,Ξ³2β,Ξ΄2β)=(r3β,Ξ³3β,Ξ΄3β)+(r4β,Ξ³4β,Ξ΄4β).
[TABLE]
plus eight non trivial suitable 3-binomials of I+β(Ξ·)3β:
[TABLE]
None of these eight non trivial suitable 3-binomials admits an I+β(Ξ·)3β-sequence from wΞ±+β to
wΞ±ββ . For instance, consider the non trivial suitable 3-binomial wΞ±=wΞ±+ββwΞ±ββ=w(0,0,0)βw(1,0,0)βw(2,0,5)ββw(1,0,1)βw(1,0,2)2β of I(X5β). Assume that {wa1β,β¦,watβ} is an I+β(Ξ·)3β-sequence from
wΞ±+ to wΞ±β. Therefore wΞ±+βwa2β is a trivial
suitable 3-binomial. So there are w(r,Ξ³,Ξ΄)ββ{w(0,0,0)β,w(1,0,0)β,w(2,0,5)β} and a non trivial suitable 2-binomial wΞ²=wΞ²+βwΞ²β such that wΞ±+βwatβ=w(r,Ξ³,Ξ΄)βwΞ² with wΞ²+ or wΞ²β being
one of the monomials w(0,0,0)βw(2,0,5)β or w(1,0,0)βw(2,0,5)β.
However all non trivial suitable 2-binomials wΞΎ of I5β verifies wΞΎ+,wΞΎββ/{w(0,0,0)βw(1,0,0)β,w(0,0,0)βw(2,0,5)β, w(1,0,0)βw(2,0,5)β}. Thus we conclude that the non trivial suitable 3-binomial w(0,0,0)βw(1,0,0)βw(2,0,5)ββw(1,0,1)βw(1,0,2)2ββ/(I+β(Ξ·))2β)=I5β (see Proposition 5.4).
Now we develop our main techniques in constructing I+β(Ξ·)nβ-sequences. Let m=βi=1nβw(riβ,Ξ³iβ,Ξ΄iβ)β be a monomial of degree nβ₯2 and let w(rijββ,Ξ³ijββ,Ξ΄ijββ)β be f variables on the support of m, where 1β€f<n. If mfβ=βj=1fβw(rijββ,Ξ³ijββ,Ξ΄ijββ)β admits a suitable f-binomial mfββmfβ²β, then mβmfβ²ββsupp(m)βsupp(mfβ))βw(riβ,Ξ³iβ,Ξ΄iβ)β is a trivial suitable n-binomial.
So determining whether a monomial admits a suitable f-binomial gives us a method to construct I+β(Ξ·)nβ-sequence from a given monomial.
Let us start analyzing whether a monomial w(r,Ξ³,Ξ΄)βw(rβ²,Ξ³β²,Ξ΄β²)β of degree 2 admits a suitable 2-monomial.
Lemma 5.10**.**
Any monomial m=w(0,0,0)βw(2,Ξ³,Ξ΄)ββk[w(r,Ξ³,Ξ΄)β] admits a special suitable 2-binomial, with the following exceptions: (Ξ³,Ξ΄)=(2kβ²+β2Οββ,β2Οββββ2Οββ) if Οξ =0, and Ξ³=0 if Ξ΅=1.
Proof.
If m admits a suitable 2-binomial mβmβ² necessary mβ²=w(1,Ξ³1β,Ξ΄1β)βw(1,Ξ³2β,Ξ΄2β)β with 0β€Ξ³iββ€kβ²,
0β€Ξ΄iββ€β2dβ3Ξ³iβββ for i=1,2, and Ξ³1β+Ξ³2β=Ξ³ and Ξ΄1β+Ξ΄2β=Ξ΄. From this follows that (2,Ξ³,Ξ΄) cannot be (2,2kβ²+1,0) in case Ο=2, (2,2kβ²,1) if Ο=1 and Ξ³=0 if Ξ΅=1.
Otherwise we set Ξ³1β:=β2Ξ³ββ and Ξ³2β:=β2Ξ³ββ.
If d is even and Ξ³1β,Ξ³2β are odd or d is odd and Ξ³1β,Ξ³2β are even we take mβ²=w(1,Ξ³1β+1,β2dβ3(Ξ³1β+1)ββ)βw(1,Ξ³2ββ1,β2dβ3(Ξ³2ββ1)ββ)β. In any other case we take mβ²=w(1,Ξ³1β,β2dβ3Ξ³1βββ)βw(1,Ξ³2β,β2dβ3Ξ³2βββ)β.
β
Lemma 5.11**.**
Suppose Ξ΅=1.
Any monomial m=w(1,0,0)βw(2,Ξ³,Ξ΄)β admits a suitable 2-binomial except for Ξ³=0,β¦,k+1 and Ξ΄=max{0,dβ2Ξ³}.
Any monomial m=w(1,Ξ³,Ξ΄)βw(2,0,d)β admits a suitable 2-binomial except Ξ³=0 and Ξ΄=0,β¦,k or Ξ³=1 and Ξ΄=kβ1.
Proof.
(i) We want to determine a monomial mβ²=w(1,Ξ³1β,Ξ΄1β)βw(2,Ξ³2β,Ξ΄2β)β such that mβmβ²βI+β(Ξ·)2β. If Ξ΄>max{0,dβ2Ξ³}, we take
(1,Ξ³1β,Ξ΄1β)=(1,0,1) and (2,Ξ³2β,Ξ΄2β)=(2,Ξ³,Ξ΄β1). Let us to consider the remainder cases (2,Ξ³,max{0,dβ2Ξ³}) with Ξ³=0,β¦,2kβ²+β2Οββ.
If Ξ³>k+1, (2,Ξ³,max{0,dβ2Ξ³})=(2,Ξ³,0) and we take (1,Ξ³1β,Ξ΄1β)=(1,1,0) and (2,Ξ³2β,Ξ΄2β)=(2,Ξ³β1,0). For 0β€Ξ³β€k+1 a monomial mβ² with
Ξ³1β+Ξ³2β=Ξ³ and Ξ΄1β+Ξ΄2β=Ξ΄ does not exist because we necessarily have Ξ³1β=i and Ξ³2β=Ξ³βi for some 0β€iβ€Ξ³, 0β€Ξ΄1ββ€β2dβ3iββ and dβ2(Ξ³βi)β€Ξ΄2ββ€β22dβ3Ξ³+3iββ which give us Ξ΄<dβ2(Ξ³βi)β€Ξ΄1β+Ξ΄2β.
The proof of (ii) is analogous and we leave it to the reader.
β
Remark 5.12**.**
(1) The monomial w(0,0,0)βw(3,d,0)β admits a non trivial suitable 2-binomial only when Ο=0. Indeed, assume that w(0,0,0)βw(3,d,0)ββw(1,Ξ³1β,Ξ΄1β)βw(2,Ξ³2β,Ξ΄2β)β is a suitable 2-binomial. Then we have
Ξ³1β+Ξ³2β=3kβ²+Ο=kβ²+2kβ²+Ο. So Ξ³1β=kβ² and Ξ³2β=2kβ²+Ο=2kβ²+β2Οββ. The last equality is achieved only when Ο=0.
(2) Suppose Ο=1. Any monomial m=w(1,kβ²,0)βw(2,Ξ³,Ξ΄)β admits a suitable 2-binomial except when Ξ³=2kβ².
Indeed, if Ξ³<2kβ² we take (r1β,Ξ³1β,Ξ΄1β)=(1,kβ²β1,Ξ΄1β) and (r2β,Ξ³2β,Ξ΄2β)=(2,Ξ³+1,Ξ΄2β) with Ξ΄=Ξ΄1β+Ξ΄2β, 0β€Ξ΄1ββ€β2dβ3kβ²+3ββ and max{0,dβ2Ξ³β2}β€Ξ΄2ββ€β22dβ3Ξ³β3ββ.
If Ξ³=2kβ², since Ξ³1β<kβ² and Ξ³2ββ€2kβ² we will never have Ξ³=Ξ³1β+Ξ³2β.
(3) Suppose Ο=2. Clearly w(1,kβ²,1)βw(2,2kβ²+1,0)β and w(1,kβ²,1)βw(2,2k,2)β if Ξ΅=0 do not
admit a suitable 2-binomial.
If dβ3Ξ³ is even and Ξ΄=22dβ3Ξ³β, we take
mβ²=w(1,kβ²β2,β2dβ3(kβ²β2)ββ)βw(2,Ξ³+2,β22dβ3(Ξ³+2)ββ)β. In any other case we take mβ²=w(1,kβ²β1,β2dβ3(kβ²β1)ββ)βw(2,Ξ³+1,β22dβ3(Ξ³+1)ββ)β.
Any monomial m=w(1,kβ²,1)βw(2,Ξ³,Ξ΄)β admits a suitable 2-binomial except: Ξ³=2kβ²+1 and (Ξ³,Ξ΄)=(2kβ²,2) when Ξ΅=0.
(4) Suppose Ο=2. Any monomial m=w(1,Ξ³,Ξ΄)βw(2,2kβ²+1,0)β admits a suitable 2-binomial except Ξ³=kβ². The proof is analogous and we left it to the reader.
Proposition 5.13**.**
Suppose Ξ΅=1. Let wΞ±=wΞ±+ββwΞ±ββ be a non-trivial 3-binomial. If wΞ±+β or wΞ±ββ is one of the following:
w(0,0,0)βw(2,0,d)βw(1,0,Ξ΄)β,Ξ΄=0,β¦,k;**
w(0,0,0)βw(2,0,d)βw(3,d,0)β* and Οξ =0;*
w(0,0,0)βw(1,0,0)βw(3,d,0)β* and Οξ =0;*
w(1,0,0)βw(2,Ξ³,dβ2Ξ³)βw(3,d,0)β,Ξ³=0,β¦,k* and w(1,0,0)βw(2,k+1,0)βw(3,d,0)β;*
then there is no an I+β(Ξ·)3β-sequence from wΞ±+β to wΞ±ββ. In particular, wΞ±β/(I+β(Ξ·)2β)=Idβ and IdβξI(Xdβ)
Proof.
Let {wa1β,β¦,watβ} be an I+β(Ξ·)3β-sequence from wΞ±+β to wΞ±ββ.
So there exist w(r,Ξ³,Ξ΄)ββk[w(r,Ξ³,Ξ΄)β] and a suitable (nβ1)-binomial wΞ±β² such that wa1ββwa2β=w(r,Ξ³,Ξ΄)βwΞ±β². This implies that we can find a monomial of degree (nβ1) on the support of wΞ±+β (respectively wΞ±ββ) admitting a suitable (nβ1)-binomial.
May we suppose that wΞ±+β belongs to the above list. From Lemmas 5.10, 4.13 and 5.11 it follows that any monomial of degree 2 that we can form from
supp(wu+β) in (i), (ii) and (iii) do not admit a non trivial suitable 2-binomial contradicting the existence of an I+β(Ξ·)3β-sequence from wΞ±+β to wΞ±ββ.
In case (iv) we only have to treat the monomials associated to
(1,0,0)+(2,Ξ³,dβ2Ξ³) for Ξ³=0,β¦,k and
(1,0,0)+(2,k+1,0). Fix Ξ³β{0,β¦,k+1} and assume that there exist (1,Ξ³1β,Ξ΄1β)>(1,0,0) and (2,Ξ³2β,Ξ΄2β)<(2,Ξ³,Ξ΄) such that Ξ³1β+Ξ³2β=Ξ³ and Ξ΄1β+Ξ΄2β=dβ2Ξ³ for Ξ³=0,β¦,k; and Ξ΄1β+Ξ΄2β=0 for Ξ³=k+1. Write Ξ³2β=Ξ³βΞ³1β, therefore Ξ΄2ββ₯Ξ΄+2Ξ³1β. From this we deduce that Ξ΄1β+Ξ΄2ββ₯Ξ΄1β+Ξ΄+2Ξ³1β and hence Ξ΄1β+2Ξ³1β must be zero, that is Ξ΄1β=0=Ξ³1β, which is a contradiction. β
Proposition 5.14**.**
Suppose Ξ΅=1.
(1) The monomials
w(0,0,0)βw(2,0,d)βw(1,0,Ξ΄)β,Ξ΄=0,β¦,kβ1;**
w(0,0,0)βw(2,0,d)βw(3,d,0)β;**
w(0,0,0)βw(1,0,0)βw(3,d,0)β;**
w(1,0,0)βw(2,Ξ³,dβ2Ξ³)βw(3,d,0)β,Ξ³=0,β¦,kβ1**
admit a suitable 3-binomial of I+β(Ξ·)3β.
(2) The monomials w(0,0,0)βw(2,0,d)βw(1,0,k)β, w(1,0,0)βw(2,k,1)βw(3,d,0)β and w(1,0,0)βw(2,k+1,0)βw(3,d,0)β do not admit a suitable 3-binomial.
Proof.
(1) It is enough to exhibit explicitly a 3-binomial in each case.
For any Ξ΄β{0,β¦,kβ1} we have w(0,0,0)βw(2,0,d)βw(1,0,Ξ΄)ββw(1,0,k)βw(1,0,k)βw(1,0,Ξ΄+1)β belong to I+β(Ξ·)3β.
We have w(0,0,0)βw(2,0,d)βw(3,d,0)ββw(1,0,k)βw(2,k,β2k+1ββ)βw(2,k+1,β2k+1ββ)ββI+β(Ξ·)3β.
We have w(0,0,0)βw(1,0,0)βw(3,d,0)ββw(1,β2kβ²+β2Οββββ,0)βw(1,β2kβ²+β2Οββββ,0)βw(2,2kβ²+β2Οββ,0)ββI+β(Ξ·)3β.
For all 0β€Ξ³β€kβ1, w(1,0,0)βw(2,Ξ³,dβ2Ξ³)βw(3,d,0)ββw(2,Ξ³+1,max{0,dβ2Ξ³β2})βw(2,k,1)+(2,k,1)β is a suitable 3-binomial.
(2) If w(0,0,0)βw(2,0,d)βw(1,0,k)ββw(1,Ξ³1β,Ξ΄1β)βw(1,Ξ³2β,Ξ΄2β)βw(1,Ξ³3β,Ξ΄3β)β is a suitable 3-binomial, we must have Ξ³1β=Ξ³2β=Ξ³3β=0 and Ξ΄1β+Ξ΄2β+Ξ΄3β=3k+1. However, Ξ΄1β,Ξ΄2β,Ξ΄3ββ€k.
If w(2,Ξ³1β,Ξ΄1β)βw(2,Ξ³2β,Ξ΄2β)βw(2,Ξ³3β,Ξ΄3β)β forms a 3-binomial, in these cases, since Ξ΄1β+Ξ΄2β+Ξ΄3ββ{0,1}, we must have Ξ³1β,Ξ³2β,Ξ³3ββ₯k. But when Ξ³=k, Ξ³1β+Ξ³2β+Ξ³3β=3k+1 implies that some (2,Ξ³iβ,Ξ΄iβ)=(2,k,1). Finally, if Ξ³=k+1, then Ξ³1β,Ξ³2β,Ξ³3ββ₯k+1 hence we find a similar argument.
β
Notice that the last two Propositions are false for even values of d. For instance we have that for d even {w(0,0,0)βw(1,0,0)βw(2,0,d)β,w(1,0,k)2βw(1,0,0)β} is a I+β(Ξ·)3β-sequence. For sake of completeness we exhibit a complete example.
Example 5.15**.**
We center in I(X4β)=I4β.
We only have to check that all monomials as in Proposition 5.14(2) contain a
submonomial of degree 2 admitting a non trivial suitable 2-binomial. Indeed,
w(0,0,0)βw(2,0,4)ββw(1,0,2)2β and w(1,0,0)βw(3,4,0)ββw(2,2,0)2β are suitable 2-binomials of I4β, from which the result follows.
In the sequel we fix nβ₯3, otherwise indicated.
Any non trivial suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ is associated to a lattice point Ξ± of the form:
[TABLE]
[TABLE]
for integers 0β€a,b,c,e,A,B,C,Eβ€n and aA=0=eE. Since wΞ± is a suitable n-binomial, we have restrictions a+b+c+e=A+B+C+E and b+2c+3e=B+2C+3E.
Proposition 5.16**.**
Let wΞ±=wΞ±+ββwΞ±ββ be a non trivial suitable
n-binomial with w(0,0,0)ββsupp(wΞ±) or w(3,d,0)ββsupp(wΞ±). Assume that wΞ±+β,wΞ±βββ/M3Οβ. Then there exist I+β(Ξ·)nβ-sequences {wΞ±+β,β¦,wΞ±+β²β} and {wΞ±ββ²β,β¦,wΞ±ββ} where w(0,0,0)β,w(3,d,0)ββ/supp(wΞ±+β²β)βͺsupp(wΞ±ββ²β).
Proof.
We write wΞ±+=a(0,0,0)+βi=1bβ(1,Ξ³i1β,Ξ΄i1β)+βj=1cβ(2,Ξ³j2β,Ξ΄j2β)+e(3,d,0) and we assume that a>0 or e>0. Analogous we deal with wΞ±ββ. It is enough to see that we can always decrease the value of a+e until we reach 0. We analyze separately several cases according to the value of d:
Case 1: Assume Ξ΅=0 and Ο=0. First we observe that the hypothesis wΞ± non-trivial implies (b,c)ξ =(0,0) or (b,c)=(0,0) and a=e. If (b,c)=(0,0) and a=e we have w(0,0,0)aβw(3,d,0)aβ=w(1,kβ²,0)aβw(2,2kβ²,0)aβ. Otherwise, since m=w(3,d,0)βw(1,Ξ³11β,Ξ΄11β)β
(resp. m=w(0,0,0)βw(2,Ξ³12β,Ξ΄12β)β) admits a special suitable 2-binomial mβmβ² with mβ²=w(2,Ξ³c+12β,Ξ΄c+12β)βw(2,Ξ³c+22β,Ξ΄c+22β)β (resp. mβ²=w(1,Ξ³b+11β,Ξ΄b+11β)βw(1,Ξ³b+21β,Ξ΄b+21β)β),
we can write
[TABLE]
[TABLE]
and build an I+β(Ξ·)nβ-sequence {wΞ±+β,wa1β} with degw(0,0,0)βwa1β+degw(3,d,0)βwa1β<a+e=degw(0,0,0)βwΞ±+β+degw(3,d,0)βwΞ±+β and we have decreased by 1 the value of a+e.
Case 2: Assume Ξ΅=0 and 1β€Οβ€2. The hypothesis wΞ± non-trivial implies (b,c)ξ =(0,0) and we can argue as in Case 1 unless
wΞ±+β=w(0,0,0)aβw(1,kβ²,0)bβw(2,2kβ²,1)cβw(3,d,0)eβ (resp. wΞ±+β=w(0,0,0)aβw(1,kβ²,1)bβw(2,2kβ²+1,0)cβw(3,d,0)eβ) but such wΞ±+β does not admit a non-trivial n-binomial wΞ±+ββwΞ±ββ.
Case 3: Assume Ξ΅=1 and Ο=0.
Since w(0,0,0)βw3,d,0)β=w(1,kβ²,0)βw(2,2kβ²,0)β we can argue as in the Case 1 unless wΞ±+β=w(0,0,0)aβw(1,0,0)bβw(2,0,d)cβ or wΞ±+β=w(1,0,0)bβw(2,0,d)cβw(3,d,0)eβ, the fact that
wΞ±+ββwΞ±ββ is non-trivial implies b,c>0 and the hypothesis wΞ±+ββ/M3Οβ implies a+b+c>3 (resp. b+c+e>3). Set m=w(0,0,0)βw(1,0,0)βw(2,0,,d)β (resp. m=w(1,0,0)βw(2,0,d)βw(3,d,0)β). By Proposition 5.14
w(0,0,0)βw(2,0,d)βw(1,0,0)ββw(1,0,k)βw(1,0,k)βw(1,0,1)β
(resp. w(1,0,0)βw(2,0,d)βw(3,d,0)ββw(2,1,dβ2})βw(2,k,1)+(2,k,1)β) and we apply the same game decreasing a (resp. e) by one.
Case 4: Assume Ξ΅=1 and 1β€Οβ€2. Notice that from the hypothesis wΞ± non trivial we have (b,c)ξ =0. So we proceed as in Case 1 unless wΞ±+=w(0,0,0)aβw(1,0,0)bβw(1,kβ²,0)cβw(2,0,d)fβw(2,2kβ²,1)gβw(3,d,0)eβ with (b,c,f,g)ξ =(0,0,0,0) (resp. wΞ±+β=w(0,0,0)aβw(1,0,0)bβw(1,kβ²,1)cβw(2,0,d)fβw(2,2kβ²+1,0)gβw(3,d,0)eβ and (b,c,f,g)ξ =(0,0,0,0)). Since wΞ±+ββ/M31β we have
(c,g)ξ =(0,0) or a+b+f+g+e>3.
By Proposition 5.14 we have
w(0,0,0)βw(1,0,0)βw(2,0,d)ββw(1,0,k)βw(1,0,k)βw(1,0,1)β,
w(0,0,0)βw(2,0,d)βw(3,d,0)ββw(1,0,k)βw(2,k,β2k+1ββ)βw(2,k+1,β2k+1ββ)β,
w(0,0,0)βw(1,0,0)βw(3,d,0)β βw(1,1,0)βw(1,kβ²,0)βw(2,2kβ²,0)β
and
w(1,0,0)βw(2,0,d)β w(3,d,0)ββw(2,1,dβ2)βw(2,k,1)βw(2,k,1)β are non trivial suitable 3-binomials (resp. w(0,0,0)βw(1,0,0)βw(2,0,d)ββw(1,0,k)βw(1,0,k)βw(1,0,2)β,
w(0,0,0)βw(2,0,d)βw(3,d,0)β βw(1,0,k)βw(2,k,β2k+1ββ)βw(2,k+1,β2k+1ββ)β,
w(0,0,0)βw(1,0,0)βw(3,d,0)ββw(1,1,0)βw(1,kβ²,0)βw(2,2kβ²+1,0)β and
w(1,0,0)βw(2,0,d)β w(3,d,0)ββw(2,1,dβ2)βw(2,k,1)βw(2,k,1)β).
Then we argue as in Case 3 decreasing a and e unless wΞ±+β=w(0,0,0)aβw(1,kβ²,0)cβw(2,2kβ²,1)gβw(3,d,0)eβ (resp. w(0,0,0)aβw(1,kβ²,1)cβ w(2,2kβ²+1,0)gβw(3,d,0)eβ)
but such monomial does not admit a non trivial suitable n-binomial and the proof is completed.
β
Remark 5.17**.**
It is easy to observe that any suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ=βi=1bβw(1,Ξ³i1β,Ξ΄i1β)ββj=1cβw(2,Ξ³j2β,Ξ΄j2β)βββi=1cβ²βw(1,Ξ³i3β,Ξ΄i3β)ββj=1bβ²βw(2,Ξ³j4β,Ξ΄j4β)β satisfies b=bβ² and c=cβ².
Example 5.18**.**
(1) Fix d=4 and consider the non trivial 3-binomial
[TABLE]
Since w(0,0,0)βw(2,0,4)ββw(1,0,2)2β is a non trivial 2-binomial, we define wa1β=w(1,0,0)βw(1,0,2)2β and we get
an I+β(Ξ·)3β-sequence {w(0,0,0)βw(1,0,0)βw(2,0,4)β,w(1,0,0)βw(1,0,2)2β,w(1,0,1)βw(1,0,1)βw(1,0,2)β} from w(0,0,0)βw(1,0,0)βw(2,0,4)β to w(1,0,1)βw(1,0,1)βw(1,0,2)β where wΞ±+β²β=wa1β.
(2) Fix d=5 and consider I(X5β) and the non trivial 4-binomial
w(0,0,0)βw(1,0,0)βw(2,0,5)βw(3,5,0)ββw(1,1,0)2βw(2,1,3)βw(2,2,2)β.
We take the suitable 3-binomial
w(0,0,0)βw(1,0,0)βw(2,0,5)ββw(1,0,1)βw(1,0,2)2β and we define
wa1β:=w(1,0,1)βw(1,0,2)2βw(3,5,0)β. We observe that w(0,0,0)ββ/supp(wa1β).
The monomial w(1,0,1)βw(3,5,0)β admits a suitable 2-binomial
w(1,0,1)βw(3,5,0)ββw(2,2,1)βw(2,3,0)β. We now define wa2β:=w(1,0,2)2βw(2,2,1)βw(2,3,0)β. We obtain the I+β(Ξ·)4β-sequence
[TABLE]
with w(0,0,0)β,w(3,5,0)ββ/supp(wa2β).
(3) Fix d=5 and consider the non trivial 4-binomial
[TABLE]
Since w(0,0,0)βw(2,2,1)ββw(1,1,0)βw(1,1,1)β and w(1,0,2)βw(3,5,0)ββw(2,2,2)βw(2,3,0)β
are suitable 2-binomials,
{w(0,0,0)βw(2,0,5)βw(2,2,1)βw(2,3,0)β,w(1,1,0)β w(1,1,1)βw(2,0,5)βw(2,3,0)β}
and {w(1,0,2)2βw(2,2,2)βw(2,3,0)β,w(1,0,2)3βw(3,5,0)β}
are the I+β(Ξ·)4β-sequences required in Proposition 5.16.
Thus is an I+β(Ξ·)4β-sequence. Furthermore, gluing them we obtain the I+β(Ξ·)4β-sequence
[TABLE]
We now analyze whether a monomial m=w(r1β,Ξ³1β,Ξ΄1β)βw(r2β,Ξ³2β,Ξ΄2β)β with r1β,r2ββ{1,2} admits a non-trivial suitable 2-binomial mβmβ² with mβ²=w(r3β,Ξ³3β,Ξ΄3β)βw(r4β,Ξ³4β,Ξ΄4β)β and r3β,r4ββ{1,2}. This problem can be reformulate as follows. For which sβ₯0, setting Ξ³3β:=Ξ³1βΒ±s and Ξ³4β:=Ξ³2ββs, there exist
max{0,(riββ1)dβ2Ξ³iβ}β€Ξ΄iββ€β2riβdβ3Ξ³iβββ, i=3,4, such that Ξ΄3β+Ξ΄4β=Ξ΄1β+Ξ΄2β.
Lemma 5.19**.**
With the above notation, there exist such Ξ΄3β and Ξ΄4β with the following exceptions.
- (1)
For any 1β€r1β,r2ββ€2, if (r1βd1ββ3Ξ³1β) and (r2βd2ββ3Ξ³2β) are even, s is odd, and
Ξ΄1β and Ξ΄2β are the maximum ones. We call it the maximum bound problem.
2. (2)
Assume r2β=2.
If r1β=1, when doing Ξ³1β+s and Ξ³2ββs
we have Ξ³2ββs<k+Ξ΅ and Ξ΄1β+Ξ΄2β<max{0,dβ2Ξ³2ββ2s}.
If r1β=2, when doing Ξ³1β+s and Ξ³2ββs we have Ξ΄1β+Ξ΄2β<max{0,dβ2Ξ³1ββ2s}+max{0,dβ2Ξ³2β+2s} and one of the following cases:
Ξ³1ββ₯k+Ξ΅* and Ξ³2ββs<k+Ξ΅,*
Ξ³1β<k+Ξ΅,Ξ³1β+sβ₯k+Ξ΅,Ξ³2ββ₯k+Ξ΅* and Ξ³1β>Ξ³2ββs,*
Ξ³1β,Ξ³2β<k+Ξ΅,Ξ³1β+s>k+Ξ΅.
We call it the minimum bound problem.
Proof.
We have
max{0,(r1ββ1)dβ2Ξ³1β}+max{0,(r2ββ1)dβ2Ξ³2β}β€Ξ΄1β+Ξ΄2ββ€β2r1βdβ3Ξ³1βββ+β2r2βdβ3Ξ³2βββ and
max{0,(r1ββ1)dβ2(Ξ³1β+s)}+max{0,(r2ββ1)dβ2(Ξ³2ββs)}β€Ξ΄3β+Ξ΄4ββ€β2r1βdβ3(Ξ³1β+s)ββ+β2r2βdβ3(Ξ³2ββs)ββ.
So the result is clear for those values
max{0,(r1ββ1)dβ2(Ξ³1β+s)}+max{0,(r2ββ1)dβ2(Ξ³2ββs)}β€Ξ΄1β+Ξ΄2ββ€β2r1βdβ3(Ξ³1β+s)ββ+β2r2βdβ3(Ξ³2ββs)ββ. Let us study the remainder cases.
(1) From the properties of the floor and ceiling functions we have
[TABLE]
[TABLE]
Furthermore β2r1βdβ3(Ξ³1β+s)ββ+β2r2βdβ3(Ξ³2ββs)ββ<β2r1βdβ3Ξ³1βββ+β2r2βdβ3Ξ³2βββ only when
(r1βdβ3Ξ³1β) and (r2βdβ3Ξ³2β) are even and s is odd. From this (1) follows immediately.
(2) is obtained determining which values max{0,(r1ββ1)dβ2Ξ³1β}+max{0,(r2ββ1)dβ2Ξ³2β}β€Ξ΄1β+Ξ΄2β<max{0,(r1ββ1)dβ2(Ξ³1β+s)}+max{0,(r2ββ1)dβ2(Ξ³2ββs)}.
β
Up to here we have proved the following. Suppose given a non trivial suitable n-binomial wΞ±=wΞ±+ββwΞ±ββ such that wΞ±+β,wΞ±βββ/M3Οβ. If w(0,0,0)ββsupp(wΞ±) or w(3,d,0)ββsupp(wΞ±), there exit I+β(Ξ·)nβ-sequences {wΞ±+β,β¦,wΞ±+β²β} and {wΞ±ββ²β,β¦,wΞ±ββ} such that w(0,0,0)β,w(3,d,0)ββ/supp(wΞ±ββ²β)βͺsupp(wΞ±ββ). Clearly wΞ±β²:=wΞ±+β²ββwΞ±ββ²ββ(I+β(Ξ·))nβ. Notice that wΞ±β² could be
trivial or even more it could be zero. In the first case {wΞ±+β,β¦,wΞ±+β²β,wΞ±ββ²β,β¦,wΞ±ββ} is
an I+β(Ξ·)nβ-sequence. In the other case let t+β,tβββ₯0 be the length of the respective I+β(Ξ·)nβ-sequences. Since wΞ± is non trivial we must have t+β>0 or tββ>0.
Assume t+β>0 (analogously, for t+β=0 and tββ>0). Therefore {wΞ±+β,β¦,wat+ββ1β,wΞ±ββ²β,β¦,wΞ±ββ} is an I+β(Ξ·)nβ-sequence.
In next Proposition we deal with the case that wΞ±+β²ββwΞ±ββ²β is neither trivial nor zero.
Proposition 5.20**.**
Let wΞ±=βi=1tβw(1,Ξ³iβ,Ξ΄iβ)ββi=t+1nβw(2,Ξ³iβ,Ξ΄iβ)βββi=1tβw(1,Ξ³iβ²β,Ξ΄iβ²β)ββi=t+1nβw(2,Ξ³iβ²β,Ξ΄iβ²β)β
be a non trivial suitable n-binomial with nβ₯3. Therefore, there exist I+β(Ξ·)nβ-sequences {wΞ±+β,β¦,wrΞ±+ββ} and {wΞ±ββ,β¦,wuΞ±βββ} with
wrΞ±+ββ=βi=1tβw(1,Ξ³i1β,Ξ΄i1β)ββi=t+1nβw(2,Ξ³i1β,Ξ΄i1β)β
and
wuΞ±βββ=βi=1tβw(1,Ξ³i2β,Ξ΄i2β)ββi=t+1nβw(2,Ξ³i2β,Ξ΄i2β)β
satisfying Ξ³i1β=Ξ³i2β for all i=1,β¦,n.
Proof.
May we assume that Ξ³1ββ₯β―β₯Ξ³tβ,Ξ³t+1ββ₯β―β₯Ξ³nβ (respectively Ξ³iβ²β) and let Ξ³ββ be first such that Ξ³jβξ =Ξ³jβ²β. May we also assume that Ξ³ββ=Ξ³ββ²β+s with
s>0. Hence βjξ =ββΞ³jβ+s=βjξ =ββΞ³jβ²β. Let Ξ³iβ be the first such that Ξ³iβ<Ξ³iβ²β with i>β and let siβ>0 be such that Ξ³iβ+siβ=Ξ³iβ²β. Now we discuss two cases.
Case 1: sβ€siβ. According to Lemma 5.19 when doing Ξ³βββs and Ξ³iβ+s the minimum bound problem (shortly, mbp) does not take place and the maximum bound problem (shortly, MBP) appears when rββdβ3Ξ³ββ, riβdβ3Ξ³iβ are even, s is odd, Ξ΄ββ=2rββdβ3Ξ³βββ andΞ΄iβ=2riβdβ3Ξ³iββ. If
MBP does not appear
we define,
[TABLE]
and then {wa+β,wa2β} is an I+β(Ξ·)nβ-sequence and wa2β,waββ share the same Ξ³ in position β.
We assume that the MBP appears and we divide the discussion in several subcases based on the parity of d.
Ξ΅=0, Ξ³ββ and Ξ³iβ even and s odd.
Ξ΅=1, rlβ=riβ=2, Ξ³ββ and Ξ³iβ even and s odd.
Ξ΅=1 , rlβ=riβ=1, Ξ³ββ and Ξ³iβ odd and s odd.
Ξ΅=1, rlβ=1, riβ=2, Ξ³ββ odd, Ξ³iβ even and s odd.
We treat 1.1, the remainder cases are similar and we leave them to the reader. We will modify both wa+β and waββ. When doing Ξ³βββ(s+1) and Ξ³iβ+(s+1) the MBP disappears.
Since Ξ³ββ and Ξ³iβ are even and s is odd we get that Ξ³ββ²β is odd. If
Ξ³iβ²β<riβkβ²+β3riβΟββ when doing Ξ³ββ²ββ1 and Ξ³iβ²β+1
the mbp does not appear and we set
[TABLE]
[TABLE]
{wΞ±+β,wa2β} and {wa2β²β,wΞ±ββ²β} are I+β(Ξ·)nβ-sequences and wa2β,wa2β²β share the same Ξ³ in position β.
If Ξ³iβ²β=riβkβ²+β3riβΟββ, there is no problem when doing Ξ³ββ²β+1, Ξ³iβ²ββ1 and set
[TABLE]
[TABLE]
In any case wa2β and wΞ±ββ (resp. wa2β²β) share the same Ξ³ in position β.
Case 2: s>siβ. Arguing as in Case 1 we distinguish cases 1.1, 1.2, 1.3 and 1.4 and we treat the first one. Assume Ξ³ββ, Ξ³iβ even, s odd and Ξ΄ββ=2rββdβ3Ξ³βββ,Ξ΄iβ=2riβdβ3Ξ³iββ. Hence now Ξ³iβ²β is odd and we can argue as in Case 1 when doing Ξ³ββ²β+1 and Ξ³iβ²ββ1. If s>siβ, then wa2β and wΞ±ββ (resp. wa2β²β) verifies the same hypothesis that wΞ±+β and wΞ±ββ but now we have Ξ³βββsiβ and Ξ³ββ²β (resp. Ξ³βββ(siββ1) and Ξ³ββ²β+1) in position β. Next we apply the same strategy to wa2β and so on until in step t>1 the resulting monomial watβ verifies Case 1.
The result follows by iterating the above argument.
β
Remark 5.21**.**
Notice that not necessarily wrΞ±+βββwuΞ±βββ
is a non trivial suitable n-binomial. In which case we obtain an Inβ-sequence from wΞ±+β to wΞ±ββ arguing as below Proposition 5.16.
Example 5.22**.**
In Example 5.18 (2), we had
w(0,0,0)βw(1,0,0)βw(2,0,5)βw(3,5,0)ββw(1,1,0)2βw(2,1,3)βw(2,2,2)β and we have build the I+β(Ξ·)4β-sequence
[TABLE]
Now we apply Proposition 5.20 to the non trivial 4-binomial
[TABLE]
We have Ξ³1β=Ξ³2β=0,Ξ³3β=2,Ξ³4β=3 and
Ξ³1β²β=Ξ³2β²β=1,Ξ³3β=1,Ξ³4β=2, with Ξ³1β=Ξ³1β²β+1. The first Ξ³iβ<Ξ³iβ²β corresponds to Ξ³3β with s3β=1. Then we choose the suitable 2-binomial w(1,1,0)βw(2,1,3β)βw(1,0,1)βw(2,2,2)β and we define wa2β:=w(1,0,1)βw(1,1,0)βw(2,2,2)2β. Note the Ξ³βs involved in
wa2β by Ξ³~βiβ, i=1,2,3,4. Now Ξ³~β1β=Ξ³1β²β,Ξ³~β2β=1, Ξ³~β3β=Ξ³~β4β=2. The first
Ξ³~βiβ>Ξ³iβ²β is Ξ³2β=Ξ³2β²β+1 and the first Ξ³jβ<Ξ³jβ²β with jβ₯3 is Ξ³4β=2 with s4β=1. Then we choose the suitable 2-binomial w(1,1,0)βw(2,2,2)ββw(1,0,2)βw(2,3,0)β and we define
wa3β:=w(1,0,1)βw(1,0,2)βw(2,2,2)βw(2,3,0)β. We have obtained an I+β(Ξ·)4β-sequence from w(0,0,0)βw(1,0,0)βw(2,0,5)βw(3,5,0)β to w(1,1,0)2βw(2,1,3)βw(2,2,2)β. Precisely, {w(0,0,0)βw(1,0,0)βw(2,0,5)βw(3,5,0)β,w(1,0,1)βw(1,0,2)2βw(3,5,0)β,w(1,0,2)2βw(2,2,1)βw(2,3,0)β,w(1,0,1)βw(1,0,2)βw(2,2,2)βw(2,3,0)β, w(1,0,1)βw(1,1,0)βw(2,2,2)2β,w(1,1,0)2βw(2,1,3)βw(2,2,2)β}.
Finally we consider wΞ±=wΞ±+ββwΞ±ββ a non trivial suitable n-binomial as in Proposition 5.20. Assume that the resulting suitable n-binomial wrΞ±+βββwuΞ±βββ=βi=1tβw(1,Ξ³i1β,Ξ΄i1β)ββi=t+1nβw(2,Ξ³i1β,Ξ΄i1β)βββi=1tβw(1,Ξ³i2β,Ξ΄i2β)ββi=t+1nβw(2,Ξ³i2β,Ξ΄i2β)β is non trivial and non zero. To prove Theorem 5.6 it is enough now to show that wrΞ±+βββwuΞ±βββ admits an
I+β(Ξ·)nβ-sequence.
wrΞ±+βββwuΞ±βββ verifies Ξ³i1β=Ξ³i2β for all 1β€iβ€n.
For each Ξ΄i1β<Ξ΄i2β, 1β€iβ€n, set aiβ=Ξ΄i2ββΞ΄i1β and biβ=0, otherwise set aiβ=0 and biβ=Ξ΄i1ββΞ΄i2β. Therefore
Ξ΄11β+a1ββb1β+β―+Ξ΄n1β+anββbnβ=Ξ΄12β+β―+Ξ΄n2β, which implies that
a1β+β―+anβ=b1β+β―+bnβ. May we assume that
a1β>0. Hence Ξ΄21β+β―+Ξ΄n1β>Ξ΄22β+β―+Ξ΄n2β. Without lost of generality we can assume that Ξ΄i1β>Ξ΄i2β, i=2,β¦,n. Then biβ>0 and Ξ΄i1β+biβ=Ξ΄i2β,
i=2,β¦,n. So a1ββ€b2β+β―+bnβ and we can consider ciββ€biβ such that a1β=c2β+β―+cnβ. Set
[TABLE]
{wrΞ±+ββ,wa2β} is an I+β(Ξ·)nβ-sequence. If Ξ΄21ββc2β=Ξ΄22β, then {wrΞ±+ββ,wa2β,wuΞ±βββ} is an I+β(Ξ·)nβ-sequence and we finish. Else inductively set
[TABLE]
Since at some point 2β€tβ€n we achieve watββwuΞ±βββ trivial, we construct an I+β(Ξ·)nβ-sequence {wrΞ±+ββ,wa2β,β¦,watβ,wuΞ±βββ} and the proof of Theorem 5.6 is completed.
β‘
6. Final remarks and open problems
In the previous sections we have explicitly described I(Xdβ). Next goal will be to compute a minimal free resolution of I(Xdβ) or at least its graded Betti numbers.
Using the program Macaulay2 we have computed a minimal free R-resolution of the ideal of the GT-threefold Xdβ for d=4,5,6 and we have got:
d=4:
[TABLE]
[TABLE]
d=5:
[TABLE]
[TABLE]
d=6:
[TABLE]
[TABLE]
[TABLE]
It follows from [16, Proposition 13] that Xdβ is arithmetically Cohen-Macaulay (see also [4] and [15]). We would like to address the following problem.
Problem 6.1**.**
Find explicitly a minimal free R-resolution of I(Xdβ) for all dβ₯4.