The weak Lefschetz property for Artinian Gorenstein algebras of codimension three

Contents lists available atScienceDirect

Journal of Pure and Applied Algebra

www.elsevier.com/locate/jpaa

The weak Lefschetz property for Artinian Gorenstein algebras of codimension three

Rosa M. Miró-Roig^a, QuangHoa Tran^b,∗

a UniversitatdeBarcelona,DepartamentdeMatemàtiquesiInformàtica,GranViadelesCorts Catalanes585,08007Barcelona,Spain

bUniversityofEducation,HueUniversity,34LeLoiSt.,HueCity, VietNam

a r t i cl e i n f o a b s t r a c t

Articlehistory:

Received30July2019

Receivedinrevisedform4December 2019

Availableonline7January2020 CommunicatedbyS.Iyengar

MSC:

Primary13E10;13H10;secondary 13A30;13C40

Keywords:

Apéryset

ArtinianGorensteinalgebras Hessians

Macaulaydualgenerators Numericalsemigroups WeakLefschetzproperty

We studythe weak Lefschetz propertyof a classof graded Artinian Gorenstein algebras of codimension three associated in a natural way to the Apéry set of a numerical semigroup generatedby four natural numbers.We show that these algebras have the weak Lefschetz property whenever the initial degree of their definingidealissmall.

©2020ElsevierB.V.Allrightsreserved.

1. Introduction

TheweakLefschetzproperty(WLPforshort)foranArtiniangradedalgebraAoverafieldKsimplysays thatthereexistsalinearformLthatinduces,foreachi,amultiplicationmap×L: [A]i−→[A]i+1thathas maximalrank,i.e.thatiseitherinjectiveorsurjective.Atfirstglancethismightseemtobeasimpleproblem of linear algebra. However, determining which graded Artinian K-algebras have the WLP is notoriously difficult. Manyauthors havestudied theproblem from manydifferent points ofview, applyingtools from representationtheory,topology,vectorbundletheory,planepartitions,splines,differentialgeometry,among others(seeforinstance[3,11,13,17,20,21]).TheroleofthecharacteristicofKinthisproblemhasalsobeen animportant,andonlysuperficially understood,aspectof thesestudies.

* Correspondingauthor.

E-mailaddresses:miro@ub.edu(R.M. Miró-Roig),tranquanghoa@hueuni.edu.vn(Q.H. Tran).

https://doi.org/10.1016/j.jpaa.2020.106305 0022-4049/©2020ElsevierB.V.Allrightsreserved.

One of the most interesting open problems in this field is whetherall codimension 3 graded Artinian Gorenstein algebras have the WLP in characteristic zero. In the special case of codimension 3 complete intersections,apositiveanswerwasobtainedincharacteristiczeroin[14] usingtheGrauert-Mülichtheorem.

For positivecharacteristic, on theother hand,only thecase ofmonomial complete intersections has been studied(see[4,6,7]),applying manydifferent approachesfromcombinatorics.

For thecase of codimension3Gorensteinalgebras thatarenot necessarilycomplete intersections,it is knownthatforeachpossibleHilbertfunctionanexampleexistshavingtheWLP[12].Somepartialresults are givenin[19] toshowthatforcertainHilbertfunctions,allsuchGorenstionalgebras havetheWLP.It was shown in[2] thatall codimension 3Artinian Gorenstein algebras of socle degreeat most6 havethe WLPincharacteristiczero.Butthegeneralcaseremainscompletely open.

In this work, we consider a class of graded Artinian Gorenstein algebras of codimension 3 built up startingfromtheApéry setofanumericalsemigroupgeneratedby4naturalnumbers.Ourgoalistostudy whether these algebras have the WLP. More precisely, we consider a numerical semigroup P generated by {a₁,a₂,a₃,a₄} ⊂N⁴ such thatgcd(a₁,a₂,a₃,a₄) = 1. TheApéry set Ap(P) of P with respect to the minimal generatorofthesemigroupisdefinedas follows

Ap(P) :={a∈P |a−a₁∈/P}={0 =ω₁< ω₂<· · ·< ω_a1}.

Notice that Ap(P) is a finite set and #Ap(P) = a1. Recall that anumerical semigroup P is said to be M-puresymmetric iffor each i = 1,. . . ,a1, ωi+ωa1−i+1 =ωa1 and ord(ωi)+ ord(ωa1−i+1)= ord(ωa1), where

ord(a) := max{ 4

i=1

λi|a= 4 i=1

λiai}

is the order of a∈ P. Therefore the Apéry set of a M-pure symmetric semigroup hasthe structure of a symmetric lattice.

LetK be afieldofcharacteristiczeroandconsiderthehomomorphism Φ :S:=K[x₁, . . . , x₄]−→K[P] :=K[t^a1, . . . , t^a4],

which sends x_i −→ t^ai. Then K[P] ∼= S/Ker(Φ) is a one dimensional ring associated to P. Now set S =S/(x1).ThenthereisonetoonecorrespondencebetweentheelementsofAp(P) andthegeneratorsof S as aK-vectorspace.Letmbe themaximalhomogeneousidealofS,define theassociatedgradedalgebra of theApéry setofP

A= gr_m(S) :=

i≥0

mⁱ mⁱ⁺¹.

ItfollowsthatAisastandardgradedArtinianK-algebra.Inthework[5],BryantprovedthatAisGorenstein ifandonlyifP isM-puresymmetric.In[10],Guerrieri showedthatifAisanArtinianGorensteinalgebra thatisnotacomplete intersection,then AisofformA=R/Iwith R=K[x,y,z] and

I= (x^a, y^b−x^αz^γ, z^c, x^a−αy^b−β, y^b−βz^c−γ)⊂R, (1.1) where 1≤α ≤a−1, 1≤β ≤ b−1, 1≤γ ≤c−1 and α+γ = b. The integers a,b,c,α,β and γ are determined bythestructureofAp(P),see[10, Section5].

Ourgoalistostudy theWLPforA.Ourmainresultisthefollowing(seeTheorems 3.7and3.15).

Theorem. Consider the ideal I as in (1.1).If one of the integers a,b and c is less than or equal to three, thenR/I hastheWLP.

2. ArtinianGorensteinalgebras

In this section, we will recall some standard notations and known facts that will be needed later in this work. Wefix K afieldof characteristic zeroand R =K[x1,. . . ,xn] a standard graded homogeneous polynomialringinnvariablesoverK.Let

A=R/I= D

i=0

[A]i

beagradedArtinianalgebra.Note thatAisfinite dimensionaloverK.

Definition 2.1.For any graded Artinian algebra A = R/I = D

i=0[A]_i, the Hilbert function of A is the function

hA:N−→N

defined byh_A(t)= dim_K[A]_t. AsA is Artinian, itsHilbert functionis equal to itsh-vector that onecan expressas asequence

h_A= (1 =h₀, h₁, h₂, h₃, . . . , h_D),

withhi =hA(i)>0 andD isthelastindex withthis property.TheintegerD iscalled thesocledegree of A.Theh-vectorh_Aissaidto besymmetricifhD−i=hi foreveryi= 0,1,. . . ,^D₂ .

Definition 2.2. [16, Proposition 2.1] A standardgraded Artinianalgebra A as aboveis Gorenstein if and onlyifhD= 1 andthemultiplicationmap

[A]i×[A]D−i−→[A]D∼=K isaperfectpairingforalli= 0,1,. . . ,^D₂ .

Itfollowsthattheh-vector ofagradedArtinianGorensteinissymmetric.

Definition2.3. AgradedArtinianK-algebraAissaidto havetheweakLefschetz property,briefly WLP,if there exists anelement L∈[A]₁ suchthatthemultiplication map×L: [A]_i −→[A]_i+1 hasmaximal rank foreachi.WealsosaythatahomogeneousidealIhastheWLPifR/IhastheWLP.

From now on, we onlyconsider astandard graded Artinian Gorenstein K-algebra. For these algebras, theWLPisdeterminedbyconsideringonlythemultiplication mapinonedegree.

Proposition 2.4. [18,Proposition 2.1] Let A be astandard graded Artinian Gorenstein K-algebra with the socledegree D andk:=^D₂ .Thenwehave:

(i) If D is odd, A has theWLP if and only if there isan element L ∈[A]₁ such that the multiplication map×L: [A]_k −→[A]_k+1 isanisomorphism.

(ii) If D iseven, A has the WLPif and only if there is an element L∈[A]₁ such that the multiplication map×L: [A]_k −→[A]_k+1 issurjective orequivalentlythe multiplication map×L: [A]_k−1 −→[A]_k is injective.

Proposition 2.5.[10, Theorem 2.1] Assume that G=D

i=0[G]i isa standard graded Artinian Gorenstein K-algebra withthesocledegree D thathas theWLP. If∈[G]1 isalinearelement, thenthequotientring

A= G

(0 :G)

isalsoastandardgradedArtinian GorensteinK-algebra.AssumethatGandA havethesamecodimension and setk:=^D₂ .Then

(i) IfD isodd,thenA has theWLP.

(ii) IfD isevenanddimK[G]k−1= dimK[G]k,thenA has theWLP.

Animportanttool neededtostudy whetheraGorensteinalgebrahastheWLPis theMacaulayinverse system, and especiallythe higherHessians. Wegive now somedefinitions andresults takenfrom apaper by Maeno andWatanabe [16] andfrom arecentpaper by Gondimand Zappalá[9]. Thegeneral factson theMacaulay’sinversesystemcanbe seenin[8].

Now weregardR asanR-moduleviatheoperation“◦”definedby R×R−→R

(x^α, x^β)−→x^α◦x^β=

x^β−α if βi≥αi,∀i= 1, . . . , n 0 otherwise

with x^α=x^α₁¹· · ·x^α_nⁿ andx^β =x^β₁¹· · ·x^β_nⁿ. ForapolynomialF∈R,Ann_R(F) denotes Ann_R(F) :={f ∈R|f ◦F = 0}

which is an ideal of R. It is called the annihilator of F. It is known that R/Ann_R(F) is an Artinian Gorenstein algebra. Furthermore, every Artinian Gorenstein algebra can be written in this form. More precisely, wehavethefollowing.

Proposition 2.6. [16,Theorem 2.1] LetI be an idealof R andA=R/I thequotientalgebra. Denote by m thehomogeneousmaximal idealofR.Then√

I=mandtheK-algebraAisGorenstein ifandonlyif there exists apolynomialF ∈R suchthat I= Ann_R(F).

The polynomialF intheaboveproposition iscalled theMacaulay dual generator of A=R/Ann_R(F).

Furthermore,ifF isahomogeneouspolynomialofdegreeD,thenR/Ann_R(F) isagradedArtinianGoren- steinalgebraofsocledegreeD.

Definition 2.7. LetF be apolynomialin R and d,k ≥1 betwo integers. AssumethatBd ={αi}^si=1 and Bk ={β_j}^t=1formrespectivelytheK-linearbasisof[A]_dand[A]_k.Wedefine themixedHessianofF asan (s×t)-matrix

Hess^d,k_B

d,Bk(F) := ( (αi·βj)◦F).

Inparticular, ifd=k, thenwedefinethed-thHessianofF asasquarematrix

Hess^d_Bd(F) := ( (αi·αj)◦F).

Noticethatthesingularity ofthese matricesis independentofthechosen basisandhencewe canwrite simplyHess^d(F) andHess^d,k(F).Basedonthesingularityof(mixed)HessiansofF,wecandeterminethe WLPofA=R/Ann_R(F).

Proposition2.8. [9] Assume thatA=R/AnnR(F)with F ∈[R]D andk:=^D₂ .Thenwehave:

(i) If D isodd, then A has the WLP if and only if the Hessian Hess^k(F) has maximal rank, i.e., it has nonzerodeterminant.

(ii) If D iseven,then Ahas theWLPif andonlyif themixedHessianHess^k−1,k(F) hasmaximal rank.

WeclosethissectionbyrecallingaresultontheWLPofcodimension3ArtinianGorenstein algebras.

Proposition 2.9. [2,Corollary 3.12] Incharacteristic zero, allcodimension3 Artinian Gorenstein algebras of socledegreeat most6have theWLP.

3. TheWLPforclassofArtinianGorenstein algebrasof codimension3

Fromnow on,letR=K[x,y,z] bethestandardgradedpolynomialringoverafieldK ofcharacteristic zeroandconsidertheideal

I= (x^a, y^b−x^αz^γ, z^c, x^a−αy^b−β, y^b−βz^c−γ)⊂R, (3.1) where1≤α≤a−1, 1≤β ≤b−1 and1≤γ≤c−1 suchthatα+γ=b.Itisclearthatb≤a+c−2 and bysymmetryofxandz, withoutloss ofgenerality,weassumethata≥c.First,wehavethefollowing.

Proposition3.1. Fix a,b,c,α,β,γ asabove.Set a= (x^a,y^b−x^αz^γ,z^c).Thenonehas:

(i) I=a: Ry^β.Therefore, R/I is an Artinian Gorenstein of codimension 3and thesocle degree of R/I isD=a+b+c−β−3.

(ii) TheMacaulay dualgenerator of R/I is

F = m i=0

x^a−1−iαy^{(i+1)b−1−β}z^c−1−iγ,

wherem:= max{j|a−1−jα≥0andc−1−jγ≥0}. (iii) Thefree resolutionof R/I is

0−→R(−a−b−c+β)

R(−a−b+β) R(−a−⊕c+β)

⊕ R(−b−c+β)

R(−a⊕−γ) R(−c⊕−α)

M

R(−a) R(−b)⊕

⊕ R(−c) R(−a−⊕γ+β) R(−c−⊕α+β)

R R/I−→0,

whereM isaskew-symmetricmatrix

M =

⎡

⎢⎢

⎢⎣

0 y^b−β 0 −x^α 0

−y^b−β 0 z^γ 0 0

0 −z^γ 0 y^β −x^a−α

x^α 0 −y^β 0 z^c−γ 0 0 x^a−α −z^c−γ 0

⎤

⎥⎥

⎥⎦.

Proof. Firstly,sinceaisacompleteintersection,I=a:y^β isGorenstein.Thisproves(i).Itisknownthat R/AnnR(F) is anArtinianGorenstein algebraof socle degreea+b+c−β−3. SinceI ⊂AnnR(F) and R/I isanArtinianGorensteinalgebra,by[15,Lemma1.1],I= AnnR(F).Theitem(ii)isproved.Finally, (iii) isimplied fromthestructure theoremof Gorensteinidealsofcodimension3andalsofrom astandard mappingconecomputation. 2

Oneoftheinterestingopen problemsis whetherallcodimension3graded ArtinianGorenstein algebras havetheWLPincharacteristiczero.Now letIbe anidealasin(3.1).Bytheaboveproposition,R/I isa gradedArtinianGorensteinalgebraofcodimension3,henceweareinterestedinstudyingtheWLPforR/I.

In the nextsubsections,we will provethat R/I hasthe WLP wheneverthe initial degreeof I is at most three.Inthepaper,wedenotebyId theidentitymatrixandbyM^tthetranspose matrixofamatrixM.

3.1. Theideal I containsaquadric

Inthissubsection,weconsider thesimplestcasewheretheidealI containsaquadric.

Thefirstcaseisb= 2,henceα=β =γ= 1.Weobtainthefollowing result.

Proposition 3.2. LetI betheideal

I= (x^a, y²−xz, z^c, x^a−1y, yz^c−1)⊂R with a≥c≥2.ThenR/I has theWLP.

Proof. The socle degree of R/I is D = a+c −2. Set k := ^D₂ = ^a+c₂⁻² . Set L = x−y+z. By Proposition2.4, itisenoughto showthat

×L: [R/I]k −→[R/I]k+1

is surjective,orequivalently [R/(I,L)]k+1= 0. Wehavethat R/(I, L)∼=K[x, z]/J,

where J = (x^a,x²+xz+z²,z^c,x^a−1z,xz^c−1). We will prove that [K[x,z]/J]k+1 = 0, or equivalently xⁱz^k+1−i ∈J forall0≤i≤k+ 1.Wedoit byinductiononi.As a≥c, hencec≤k+ 1.It followsthat z^k+1 andxz^k belongto J. Foranyi≥2,onehas

xⁱz^k+1−i=x²xⁱ⁻²z^k+1−i =−(z²+xz)xⁱ⁻²z^k+1−i=−xⁱ⁻²z^k+3−i−xⁱ⁻¹z^k+2−i∈J, bytheinductionhypothesis. 2

We now study thecase a = 2 or c = 2. Bysymmetry of x and z, WLOG, we canassume a≥ c = 2.

Therefore γ= 1 anda≥α+ 1=b.Moreprecisely,weconsidertheideal I_β= (x^a, y^b−x^b−1z, z², x^a−b+1y^b−β, y^b−βz)⊂R,

with1≤β≤b−1 anda≥b.SetA_β=R/I_β.ByProposition3.1, thefreeresolution ofA_β is

0 R(−a−b−2 +β)

R(−a−2 +β)

⊕ R(−b−2 +β) R(−a−⊕b+β) R(−a⊕−1)

⊕ R(−b−1)

R(−2)

⊕ R(−a) R(−b)⊕ R(−a−⊕1 +β)

⊕ R(−b−1 +β)

R Aβ 0 . (3.2)

Sincewehavethefreeresolution(3.2) ofAβ,forany integerj≥2,weget HAβ(j) =2j+ 1−

j−a+ 1 1

−

j−b+ 1 1

−

j−a+β 1

−

j−b+β 1

(3.3) +

j−a−b+β+ 1 1

+

j−a−b+β 1

,

with convention _n

m

= 0 if n < m. ByProposition3.1, thesocle degree ofA_β isD =a+b−β−1. Set k:=^D₂ .Thenk−a<0 andk−a−b+β <0,itfollowsfrom(3.3) that

H_Aβ(k) = 2k+ 1−

k−b+ 1 1

−

k−a+β 1

−

k−b+β 1

. (3.4)

TheHilbertfunctionofA_β indegreekisdetermined asfollows.

Lemma3.3. Forevery 1≤β ≤b−1,onehas HAβ(k) =

2b−β if β ≤a−b a+b−2β+ 1 if β ≥a−b+ 1.

Furthermore,if 1≤β ≤a−b−1,then

HAβ(k) =HAβ(k−1).

Proof. Firstly,weconsiderthecasewherea+b−β iseven.Hencek=^a+b−β₂ −1.Itfollowsfrom(3.4) that H_Aβ(k) =a+b−β−1−

_a−b−β

2

1

−

_a−b+β

2 −1

1

−

_b−a+β

2 −1

1

.

Sincea≥b≥β+ 1≥2.Weconsider thefollowingcases.

Case 1:a=b.Inthiscase,β hasto beevenanditiseasyto showthat H_Aβ(k) =a+b−2β+ 1.

Case 2:a=b+ 1.Inthiscase,β mustbe odd.Therefore HAβ(k) =a+b−β−1−

β+1 2 −1

1

− β−1

2 −1 1

=

2b−β if β = 1 a+b−2β+ 1 if β ≥3.

Case 3:a=b+ 2. Inthis case,β mustbeeven, henceβ ≥2.It followsthat H_Aβ(k) =a+b−β−1−

_β+2

2 −1 1

− _β−2

2 −1 1

=

2b−β if β= 2 a+b−2β+ 1 if β≥4.

Case 4:a≥b+ 3. Thena−b+β≥4.Therefore,ifβ≥a−b+ 4,then HAβ(k) =a+b−β−1−a−b+β

2 + 1−b−a+β

2 + 1

=a+b−2β+ 1.

If β≤a−b−2,then

H_Aβ(k) =a+b−β−1−a−b−β

2 −a−b+β

2 + 1

= 2b−β.

Thusweonlyconsiderthecaseβ=a−b+i, −1≤i≤3.Buta+b−β iseven,thereforeboth βanda−b are eitherevenor odd.Itfollows thatwe onlyconsider thetwocaseswhere β=a−b+ 2 orβ =a−b. If β =a−b+ 2,then astraightforwardcomputationshowsthat

HAβ(k) =a+b−2β+ 1.

Similarly,ifβ=a−b then

H_Aβ(k) = 3b−a= 2b−β.

Thuswe concludethat

HAβ(k) =

2b−β if β ≤a−b a+b−2β+ 1 if β ≥a−b+ 2 as desired.

Secondly,weconsider thecasewhere a+b−β isodd.Hencek= ^a+b−₂^β−1. Itfollowsfrom (3.4) that HAβ(k) =a+b−β−

_a−b−β+1

2

1

−

_a−b+β−1

2

1

−

_b−a+β−1

2

1

.

Since a≥b≥β+ 1≥2.Weconsiderthefollowing caseswhere a=b, a=b+ 1 ora≥b+ 2.Theproofis similar asabove(evenmoresimple).

Finally,if1≤β≤a−b−1,then

H_Aβ(k) = 2b−β.

Notice thatk−a+β <0 sincea−b≥β+ 1.It followsfrom(3.3) that H_Aβ(k−1) = 2k−1−

k−b 1

−

k−b+β−1 1

.

Ifa+b+β isodd,then

HAβ(k−1) =a+b−β−2−

a−b−β−1 2

1

−

a−b+β−1

2 −1

1

=

⎧⎪

⎪⎨

⎪⎪

⎩

2b−β if a−b=β+ 1 2b−β if a−b=β+ 3 2b−β if a−b≥β+ 5

= 2b−β.

Ifa+b+β iseven, then

H_Aβ(k−1) =a+b−β−3−

a−b−β

2 −1

1

−

a−b+β

2 −2

1

=

2b−β if a−b=β+ 2 2b−β if a−b≥β+ 4

= 2b−β.

Thusthelemmaiscompletely proved. 2

Lemma3.4. Set G=R/(x^a,y^b−x^b−1z,z²)andk=^a+b₂⁻¹ .Ifa≥b,then

HG(k) =

2b if a≥b+ 1 2b−1 if a=b.

Furthermore,if a≥b+ 3,then

HG(k) =HG(k−1).

Proof. Since GisresolvedbytheKoszulcomplexandk−a<0,wehave HG(k) =

k+ 2 2

− k

2

−

k−b+ 2 2

− k−b

2

= 2k+ 1−

k−b+ 1 1

− k−b

1

.

Ifa+b isodd,thenk= ^a+b−1₂ .Asimplecomputationshowsthat

HG(k) =

a+b−^a−₂^b−1−1−^a−₂^b−1 if a−b≥3

a+b−1 if a−b= 1

= 2b.

Ifa+b iseven,thenk=^a+b₂ −1.Itfollows that

HG(k) =

⎧⎪

⎪⎨

⎪⎪

⎩

a+b−1−^a−₂^b−^a−₂^b+ 1 if a−b≥4 a+b−1−1 if a−b= 2

a+b−1 if a=b

=

2b if a−b≥2 2b−1 if a=b.

Analogously wecancheck that

H_G(k−1) = 2k−1− k−b

1

−

k−b−1 1

=

⎧⎪

⎪⎨

⎪⎪

⎩

2b if a−b≥3 2b−1 if 1≤a−b≤2 2b−3 if a=b.

Thus, ifa−b≥3,thenH_G(k)=H_G(k−1). 2 Proposition 3.5. Assumeb≤a≤2b−3.Thentheideal

Iβ= (x^a, y^b−x^b−1z, z², x^a−b+1y^b−β, y^b−βz)⊂R has theWLP, whenevera−b+ 2≤β≤b−1.

Proof. SetA_β=R/I_β.ByProposition3.1,onehas A_β= Aβ−1

(0 :_Aβ−1y),

forall2≤β≤b−1.Notice thatthesocledegreeofAβ isD=a+b−β−1.Henceifβ=a−b+ 2,then D= 2b−3 isodd.ToprovethatAβ hastheWLPforeverya−b+ 2≤β≤b−1,byProposition2.5(i),it is enoughto provethatA_β hastheWLPwheneverD isodd.

Now letβ be aninteger such thata−b+ 2 ≤β ≤ b−1 anda+b−β is even. In this case, onehas k= ^a+b₂^−β−1.Itfollows fromLemma3.3that

H_Aβ(k) =a+b−2β+ 1.

Clearly, sinceβ≥a−b+ 2,k < b≤a.Therefore, wecantakeaK-linearbasisB=B1 B2 B3 of[A_β]_k with

B1={ui=x^k+1−iyⁱ⁻¹|i= 1,2, . . . , b−β} B2={v_i =xⁱ⁻¹y^k+1−i |i= 1,2, . . . , a−b+ 1} B3={w_i=x^k−iyⁱ⁻¹z|i= 1,2, . . . , b−β}. Ontheother hand,theMacaulaydualgeneratorofAβis

F =x^a−1y^b−β−1z+x^a−by^2b−β−1.

Toprovetheproposition,byProposition2.8,itisenoughtoshowthatHess^k_B(F) hasnonzerodeterminant.

Write

Hess^k_B(F) =

⎡

⎢⎢

⎢⎢

⎢⎢

⎣

A ... B ... C

· · · · B^t ... U ... V

· · · · C^t ... V^t ... W

⎤

⎥⎥

⎥⎥

⎥⎥

⎦ ,

where A= ( (u_i·u_j)◦F),B = ( (u_i·v_j)◦F),C = ( (u_i·w_j)◦F),U = ( (v_i·v_j)◦F),V = ( (v_i·w_j)◦F) andW = ( (wi·wj)◦F).

Notice thatA,C,U and W are thesquare matrices. It follows that thediagonal of Hess^k_B(F) from the toprightto thebottomleftcornerisequalto thediagonalsofC,U and C^t.Wewillshow thattheentries onthisdiagonalarenonzeroandtheentriesunderthislinearezero.

Indeed,astraightforwardcomputationshowsthatthematrixU = (ui,j) isasquarematrixofsizea−b+1 with

ui,j= (vi·vj)◦F = (x^i+j−2y^2k+2−i−j)◦F

=

y if i+j=a−b+ 2 0 if i+j≥a−b+ 3

sincei+j−2≤2a−2b≤2a−(a+ 3)=a−3.Similarly,thematrixC= (ci,j) isasquarematrixofsize b−β with

c_i,j= (u_i·w_j)◦F =

(x^2k−b+βy^b−β−1z)◦F if i+j =b−β+ 1 (x^2k+1−i−jy^i+j−2z)◦F if i+j ≥b−β+ 2

=

x if i+j=b−β+ 1 0 if i+j≥b−β+ 2.

ItiseasytoseethatW = ( (wi·wj)◦F) = 0 becausew_i·w_j containsz².Finally,V = (v_i,j) isamatrix ofsize (a−b+ 1)×(b−β) with

vi,j= (vi·wj)◦F = (xⁱ⁻¹y^k+1−i·x^k−jy^j−1z)◦F

= (x^k+i−j−1y^k−i+jz)◦F.

Noticethatk > b−β−1.Henceifi≤j,thenk−i+j > b−β−1.Thus(vi·wj)◦F = 0.Ifi> j,thenput :=i−j,hence1≤≤a−b≤β−2.Inthiscase,wewill seethatk−i+j > b−β−1.Indeed,onehas

k−i+j > b−β−1⇔2k−2 >2b−2β−2, wherethelast inequalityfollows fromthefactthat

2k−2≥a+b−β−2−(a−b)−(β−2)≥2b−2β.

Thus,wesee thatV = 0.

WethusconcludethattheHessianofF is

Hess^k_B(F) =

⎡

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎣

∗ · · · ∗ ∗ · · · ∗ ∗ · · · x ... · · · ... ... · · · ... ... · · · ...

∗ · · · ∗ ∗ · · · ∗ x · · · 0

∗ · · · ∗ ∗ · · · y 0 · · · 0 ... · · · ... ... · · · ... ... · · · ...

∗ · · · ∗ y · · · 0 0 · · · 0

∗ · · · x 0 · · · 0 0 · · · 0 ... · · · ... ... · · · ... ... · · · ... x · · · 0 0 · · · 0 0 · · · 0

⎤

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎦

whichhasnonzerodeterminant. 2

Proposition 3.6. Assumea≥b≥2.Thentheideal

I_β= (x^a, y^b−x^b−1z, z², x^a−b+1y^b−β, y^b−βz)⊂R has theWLP, whenever1≤β≤min{a−b+ 1,b−1}.

Proof. SetA_β=R/I_β andG=R/(x^a,y^b−x^b−1z,z²).ByProposition3.1,onehas A1= G

(0 :Gy) and Aβ= A_β−1 (0 :A_β−1y),

forall2≤β≤b−1.Noticethatifβ =a−b≤b−1,thenthesocledegreeofAβ isodd.ByProposition2.5 andLemma3.3, itisenoughto showthatA₁hastheWLP.Weconsiderthefollowingcases:

Case 1:a+b is even.ThenthesocledegreeofGisa+b−1 whichisodd.SinceGisanArtiniancomplete intersection algebra of codimension3, by [14, Corollary 2.4],G has theWLP. It follows thatA₁ has the WLPbyProposition2.5(i).

Case 2:a+b is odd.If a≥b+ 3, thenA1 hastheWLPby Proposition2.5(ii) andLemma3.4. It follows thattheremaincaseiswherea=b+ 1.Moreprecisely, wehaveto showthattheideal

I= (x^a, y^a−1−x^a−2z, z², x²y^a−2, y^a−2z)

hastheWLP.ThesocledegreeofR/I is2a−3,hencek=a−2.ByLemma3.3, onehas H_R/I(a−2) = 2a−3.

Therefore, wecansee thataK-linearbasisof[R/I]_a−2 isB=B1 B2,where B1={ui =x^a−1−iyⁱ⁻¹|i= 1,2, . . . , a−1} B2={ua+i=x^a−3−iyⁱz|i= 0,1, . . . , a−3}. Ontheother hand,theMacaulaydualgeneratorofR/I is

F =x^a−1y^a−3z+xy^2a−4.

To provethat R/I hasthe WLP, by Proposition 2.8, it is enough to show thatHess^a−2_B (F) has nonzero determinant.Write

Hess^a−2_B (F) = [Mi,j], whereMi,j= ( (ui·uj)◦F) forall1≤i,j≤2a−3.

NoticethatM isasquare matrixofsize 2a−3.First,wehave a_i,2a−2−i= (u_i·u_2a−2−i)◦F

=

⎧⎪

⎪⎨

⎪⎪

⎩

(x^a−2y^a−3z)◦F if 1≤i≤a−2 (y^2a−4)◦F if i=a−1 (x^a−2y^a−3z)◦F if a≤i≤2a−3

=x.

Nowweassumethati+j≥2a−1.If1≤i≤a−1 thenj≥a,hence a_i,j= (u_i·u_j)◦F

= (x^{3a−i−j−4}y^i+j−a−1z)◦F

= 0

sincei+j−a−1≥a−2.BythesymmetryofMi,j,weonlyconsiderthecasewherei,j≥a.Inthiscase, wehavea_i,j= 0 becauseu_i·u_j containsz².

Insummary,theHessianofF is

Hess^k_B(F) =

⎡

⎢⎢

⎢⎢

⎣

∗ ∗ · · · ∗ x

∗ ∗ · · · x 0 ... ... · · · ... ...

∗ x · · · 0 0 x 0 · · · 0 0

⎤

⎥⎥

⎥⎥

⎦

whichhasnonzerodeterminant. 2 Wenowstateourfirstmainresult.

Theorem3.7. Consider theideal I asin (3.1).If one ofthe integers a,b andc isequalto2, then R/I has theWLP.

Proof. Thetheoremwasprovedforthecaseb= 2 inProposition3.2.Bythesymmetryofxandz,wecan always assume thata ≥c. Now ifc = 2,then it follows from Propositions 3.5 and 3.6 thatR/I has the WLP. 2

3.2. The idealI contains acubic

Inthissubsection,weconsiderthecasewheretheidealI containsacubic.

Thefirstcaseweconsiderisb= 3.Denote

Iβ = (x^a, y³−x^αz^3−α, z^c, x^a−αy^3−β, y^3−βz^c+α−3)⊂R,

with a≥cand 1≤α,β ≤2 suchthatc+α≥4.Thesocle degreeofR/I_β isD =a+c−β andthe free resolutionofR/I_β is

0 R(−a−c−3 +β)

R(−a−c+β) R(−a−⊕3 +β)

⊕ R(−c−3 +β)

⊕ R(−a−3 +α)

R(−c⊕−α)

R(−3) R(−a)⊕

⊕ R(−c)

⊕ R(−a−3 +α+β)

R(−c−⊕α+β)

R R/Iβ−→0 . (3.5)

Lemma 3.8.Set G=R/(x^a,y³−x^αz^3−α,z^c)and k=^a+c₂ .If a≥c≥2,then

HG(k) =

⎧⎪

⎪⎨

⎪⎪

⎩

3c−2 if a=c 3c−1 if a=c+ 1 3c if a≥c+ 2.

Furthermore, ifa≥c+ 4,then

HG(k) =HG(k−1).

Proof. SinceGisresolvedbytheKoszulcomplexandk−a≤0,weget HG(k) =

k+ 2 2

− k−1

2

−

k−a+ 2 2

−

k−c+ 2 2

+

k−c−1 2

= 3k−

k−a+ 2 2

−

k−c+ 2 2

+

k−c−1 2

.

If a+cisodd,thenk=^a+c₂⁻¹.Itfollowsthat

H_G(k) =

3c−1 if a=c+ 1 3c if a≥c+ 3.

If a+ciseven, thenk= ^a+c₂ .Therefore

HG(k) =

3c−2 if a=c 3c if a≥c+ 2.

Analogously wecancheck that

H_G(k−1) = 3(k−1)−

k−c+ 1 2

+

k−c−2 2

=

⎧⎪

⎪⎨

⎪⎪

⎩

3c−3 if a=cora=c+ 1 3c−1 if a=c+ 2 ora=c+ 3 3c if a≥c+ 4.

Thus, ifa≥c+ 4,thenHG(k)=HG(k−1). 2

Lemma 3.9.Assume β= 1 andk=^a+c₂⁻¹ .Ifa≥c,then

H_R/I1(k) =

3c−3 if a=c 3c−3 +α if a≥c+ 1.

Furthermore,if a≥c+ 3,then

H_R/I1(k) =H_R/I1(k−1).

Proof. As k−a<0 andc≥γ+ 1≥2,itfollows from(3.5) withβ= 1 that H_R/I1(k) = 3k−

k−c+ 1 1

− k−c

1

−

k−c−α+ 2 1

.

Byconsideringthecaseswherea=c+j foreachj ∈ {0,1,. . . ,4}or a≥c+ 5,itiseasytosee that

H_R/I1(k) =

3c−3 if a=c 3c−3 +α if a≥c+ 1.

Astraightforwardcomputationsalso showsthat

HR/I1(k−1) =

⎧⎪

⎪⎨

⎪⎪

⎩

3c−6 if a=c

3c−3 if a=c+ 1 ora=c+ 2 3c−3 +α if a≥c+ 3.

Thusifa≥c+ 3 thenH_R/I1(k)=H_R/I1(k−1). 2

Thefollowing is useful to provethenext results.Recall thatthe determinantof blockmatrices canbe computedasfollows:supposeA,B,CandDarematricesofsizen×n,n×m,m×n,andm×m,respectively.

Then

det

⎡

⎢⎢

⎣

A ... B

· · · · C ... D

⎤

⎥⎥

⎦=

det(A) det(D−CA⁻¹B) ifAis invertible

det(D) det(A−BD⁻¹C) ifDis invertible. (3.6) Proposition3.10. Considertheideal

Iβ = (x^a, y³−x^αz^3−α, z^c, x^a−αy^3−β, y^3−βz^c+α−3), with a≥cand 1≤α,β ≤2suchthatc+α≥4.ThenR/Iβ has theWLP.

Proof. Recall that G=R/(x^a,y³−x^αz^3−α,z^c) isa complete intersectionof codimension3, henceit has theWLP.Weconsiderthefollowingtwo cases:

Case 1:a+c is odd. In this case, the socle degree of G is odd. Hence, R/I1 has the WLP by Proposi- tions2.5(i)and3.1. Nowifa≥c+ 3,thenR/I2 alsohastheWLPbyLemma3.9andProposition2.5(ii).

Thus,weonlyneedto provethat

I₂= (x^a, y³−x^αz^3−α, z^a−1, x^a−αy, yz^a+α−4) hastheWLP.Inthis case,onehasD= 2a−3 andk=a−2.

Subcase 1:α= 1.SetL=x−y+z.ByProposition2.4,itsufficestoshowthat

×L: [R/I₂]_a−2−→[R/I₂]_a−1

is anisomorphism,or equivalently[R/(I₂,L)]_a−1= 0. Wehave R/(I₂, L)∼=K[x, z]/J,

where J = (x^a,x³+ 3x²z+ 2xz²+z³,z^a−1,x^a−1z,xz^a−3+z^a−2).Wewill provethat[K[x,z]/J]_a−1 = 0, or equivalent xⁱz^a−1−i ∈J forall0≤i ≤a−1.We doitby inductiononi. Wefirst see thatxz^a−2 and x²z^a−3∈J sincez^a−1,xz^a−3+z^a−2∈J.Forany i≥3,onehas

xⁱz^a−1−i=x³xⁱ⁻³z^a−1−i=−(z³+ 2xz²+ 3x²z)xⁱ⁻³z^a−1−i

=−xⁱ⁻³z^a+2−i−3xⁱ⁻²z^a+1−i−3xⁱ⁻¹z^a−i∈J, bytheinductionhypothesis.

Subcase 2:α= 2.Inthiscase,

I2= (x^a, y³−x²z, z^a−1, x^a−2y, yz^a−2).

It followsfrom(3.5) that

HR/I2(a−2) =HR/I2(a−1) = a

2

− a−3

2

= 3a−6.

A K-linearbasisof[R/I₂]_a−2 isB=B1 B2 B3,where

B1={u_i=x^a−1−izⁱ⁻¹|i= 1,2, . . . , a−1} B2={v_i=x^a−2−iyzⁱ⁻¹|i= 1,2, . . . , a−2} B3={wi =x^a−3−iy²zⁱ⁻¹|i= 1,2, . . . , a−3} and aK-linearbasisof[R/I2]_a−1 isB =B1 B2 B3,where

B1 ={u_i=x^a−izⁱ⁻¹|i= 1,2, . . . , a−1} B2 ={v_i=x^a−2−iy²zⁱ⁻¹|i= 1,2, . . . , a−2} B3 ={w_i=x^a−2−iyzⁱ |i= 1,2, . . . , a−3}. SetL=x+y+z.ByProposition2.4,itisenoughtoshowthat

×L: [R/I2]a−2−→[R/I2]a−1

is an isomorphism, or equivalently the matrix representation M of ×L with respect to these bases has nonzerodeterminant.Itis easytoseethat

×L(ui) =x^a−izⁱ⁻¹+x^a−1−izⁱ+x^a−1−iyzⁱ⁻¹

×L(v_i) =x^a−1−iyzⁱ⁻¹+x^a−2−iyzⁱ+x^a−2−iy²zⁱ⁻¹

×L(wi) =x^a−1−izⁱ+x^a−2−iy²zⁱ⁻¹+x^a−3−iy²zⁱ sincey³=x²z inR/I₂.Itfollowsthat