On semigroup rings with decreasing Hilbert function

0.23 MB·

by  Anna Oneto

On semigroup rings with decreasing Hilbert function.
AnnaOneto1, Grazia Tamone2.
Dima-UniversityofGenova,viaDodecaneso35,I-16146Genova,Italy.
E-mail:[email protected], [email protected]
Abstract
6
1 InthispaperwestudytheHilbertfunctionHR ofone-dimensional semigroupringsR = k[[S]]. Forsomeclasses
ofsemigroups,bymeansofthenotionofsupportoftheelementsinS,wegiveconditionsonthegeneratorsofS in
0
2 ordertohavedecreasingHR.Whentheembeddingdimensionvandthemultiplicityeverifyv+3≤e≤v+4,the
decreaseofHRgivesexplicitdescriptionoftheApe´rysetofS.Inparticularfore=v+3,weclassifythesemigroups
n withe=13andHRdecreasing,furtherweshowthatHRisnon-decreasingife<12. FinallywededucethatHRis
a
non-decreasingforeveryGorensteinsemigroupringwithe≤v+4.
J

1
3 Keywords: Numericalsemigroup,Monomialcurve,Hilbertfunction,Ape´ryset.
MathematicsSubjectClassification: Primary:14H20;Secondary:13H10.

] 0 Introduction.
C
A
Givenalocalnoetherianring(R,m,k)andtheassociatedgradedringG = ⊕ mn/mn+1 ,aclassicalhardtopic
. n≥0
h incommutativealgebraisthestudyoftheHilbertfunctionH ofG, definedasH(cid:0) (n) = di(cid:1)m mn/mn+1 : when
R R k
t
a R is the local ring of a k-scheme X at a pointP, HR givesimportantgeometricinformation. If(cid:0)depth(G)(cid:1)is large
m enough,thisfunctioncanbecomputedbymeansoftheHilbertfunctionofalowerdimensionalring,butingeneralG
[ isnotCohen-Macaulay,evenifRhasthisproperty.
ForaCohen-Macaulayone-dimensionallocalringR,itiswellknownthatHR isanondecreasingfunctionwhen
1
GisCohen-Macaulay,butwecanhavedepth(G)=0andinthiscaseH canbedecreasing,i.e.H (n)<H (n−1)
v R R R
forsomen(see,forexample[8], [10], [11]). ThisfactcannothappenifRverifieseitherv ≤ 3, or v ≤ e ≤ v+2,
7
2 where e and v denote respectively the multiplicity and the embedding dimension of R (see [6], [7], [16]). If
3 R = k[[S]]isa semigroupring,manyauthorsprovedthatH isnon-decreasinginseveralcases: •S isgenerated
R
0 byanalmostarithmeticsequence(ifthesequenceisarithmetic,thenGisCohen-Macaulay)[17],[13] •S belongsto
0
particularsubclassesoffour-generatedsemigroups,whicharesymmetric [1],orwhichhaveBuchsbaumtangentcone
.
2 [3] •S isbalanced [12], [3] •S isobtainedbytechniquesofgluingnumericalsemigroups [2], [9] •S satisfies
0
certainconditionsonparticularsubsetsofS(seebelow)[5,Theorem2.3,Corollary2.4,Corollary2.11]. Ife≥v+3,
6
thefunctionH canbedecreasing,asshowninseveralexamples:thefirstone(withe=v+3)isin[14](hererecalled
1 R
: inExample1.6). WhenGisnotCohen-Macaulay,ausefulmethodtodescribeHRcanbefoundinsomerecentpapers
v (see[12],[3],[5]):itisbasedonthestudyofcertainsubsetsofS,calledD andC , (k ∈N).
i k k
X TheaimofthispaperisthestudyofsemigroupringsR = k[[S]]withH decreasing. Tothisgoalweintroduce
R
r andusethenotionofsupportoftheelementsinS (1.3.4);bymeansofthistoolwefirstdevelopatechnicalanalysis
a ofthesubsetsD , C inSection2. Throughthismachinery,undersuitableassumptionsontheApe´rysetofS,we
k k
prove(Section3)necessaryconditionsonS inordertohavedecreasingHilbertfunction,see(3.4),(3.6).
InSection4weapplytheseresultstothesemigroupswithv ∈{e−3,e−4}.
Forv =e−3weshowthatthedecreaseofH ischaracterisedbyaparticularstructureofthesetsC , D , C and
R 2 2 3
thatH doesnotdecreasefore ≤ 12,see(4.2),(4.3);inparticular,fore = 13,weidentifypreciselythesemigroups
R
withH decreasing,see(4.6)andexamples(4.7).
R
Incasev =e−4weobtainanalogousinformationsonthestructureofC ,C ,D ,D ,see(4.9)and(4.10).
2 3 2 3
SuchmethodsallowtoconstructvariousexamplesofsemigroupringswithdecreasingH ,see,forexample(3.2),
R
(3.7)wheree−7≤v ≤e−3;inparticularexample(3.7.1)describesasemigroupwhoseHilbertfunctiondecreases
attwodifferentlevels. TheexampleshavebeenperformedbyusingtheprogramCoCoAtogetherwithFreeMatand
Excel.
As a consequenceof some of the abovefacts, one can see thatthe semigroupsS with |C | = 3 and |C ∩Ape´ry
2 3
set| ≤ 1 cannot be symmetric. It follows that every Gorenstein ring k[[S]] with v ≥ e − 4 has non-decreasing
Hilbertfunction(4.11). ThisresultisapartialanswertotheconjecturesettledbyM.E.Rossi[15,Problem4.9]thata
Gorenstein1-dimensionallocalringhasnon-decreasingHilbertfunction.
1
1 Preliminaries.
WebreaflyrecallthedefinitionoftheHilbertfunctionforlocalrings.Let(R,m,k)beanoetherianlocald-dimensional
ring,theassociatedgradedringofRwithrespecttomisG:= mn/mn+1: theHilbertfunctionH :G−→N
n≥0 R
ofRisdefinedbyH (n) = dim (mn/mn+1). ThisfunctioniLscalled non-decreasing ifH (n−1) ≤ H (n)for
R k R R
eachn∈Nand decreasing ifthereexistsℓ∈NsuchthatH (ℓ−1)>H (ℓ), wesayH decreasingatlevelℓ.
R R R
1.1 Let R be a one-dimensional Cohen-Macaulay local ring and assume k = R/m infinite. Then there exists a
superficialelementx∈m,ofdegree1,(i.e.suchthatxmn =mn+1forn>>0).
Itiswell-knownthat
• GisCohen-Macaulay ⇐⇒ theimagex∗ ofxinGisanon-zerodivisor.
• IfGisCohen-Macaulay,thenH isnon-decreasing.
R
Webeginbysettingthenotationofthepaperandbyrecallingsomeknownusefulfacts.
Setting1.2 InthispaperRdenotesa1-dimensionalnumericalsemigroupring,i.e. R=k[[S]],wherekisaninfinite
fieldandS ={ a n , a ,n ∈N}isanumericalsemigroup ofmultiplicityeandembeddingdimensionvminimally
i i i i
generatedby{eP:= n1,n2,...,nv},with0 < n1 < ··· < nv, GCD(n1,...,nv) = 1. ThenRisthecompletionof
thelocalring k[x ,...,x ] ofthemonomialcurveC parametrizedbyx := tni (1 ≤ i ≤ v). Themaximal
1 v (x1,...,xv) i
idealofRism=(tn1,··· ,tnv)andx =teisasuperficialelementofdegree1.Letv:k((t))−→Z∪{∞}denote
1
theusualvaluation.
1. M :=S\{0}=v(m), hM =v(mh), foreachh≥1 andtheHilbertfunctionH verifies
R
H (0)=1, H (h)=|hM \(h+1)M|, foreachh≥1.
R R
2. Letg ∈S,the order ofgisdefinedas ord(g):=max{h|g ∈hM}.
3. Ap=Ape´ry(S):= {s ∈ S|s−e ∈/ S}is theApe´ry setofS withrespecttothe multiplicitye , |Ap| = e and
e+f isthegreatestelementinAp,wheref :=max{x∈IN\S}istheFrobeniusnumberofS.
Letd:=max{ord(σ)| σ ∈Ap}.
DenotebyAp :={s∈Ap|ord(s)=k}, k ∈[1,d]thesubsetoftheelementsoforderkinApe´ry.
k
4. LetR′ := R/teR,theHilbertfunctionofR′ is HR′ = [1,a1,...,ad]withak = |Apk| foreachk ∈ [1,d],
see,forexample,[12,Lemma1.3].
5. AsemigroupS iscalledsymmetricifforeachs∈S s∈Ap⇐⇒e+f −s∈Ap.
By(1.2.1)and(1.1),iford(s+e)=ord(s)+1foreachs∈S,thenGisCohen-MacaulayandH isnon-decreasing.
R
Thereforein order to focus on the possible decreasing Hilbert functions, it is useful to define the following subsets
D ,C ⊆S,wealsointroducethenotionofsupport forabetterunderstandingofthesesets.
k k
Setting1.3 1. D :={s∈S|ord(s)=k−1andord(s+e)>k}. D =D =∅, ∀k ≥r[12,Lemma1.5.2] .
k 1 k
(cid:0) (cid:1)
Dt :={s∈D suchthat ord(s+e)=t} andlet k :=min{k suchthat D 6=∅}.
k k 0 k
2. C :={s∈S|ord(s)=k and s−e∈/ (k−1)M}, i.e.,C =Ap ∪ (Dk+e), with 2≤h≤k−1 .
k k k h h
S (cid:8) (cid:9)
C ={n ,··· ,n }, C =Ap , C =(D3+e)∪Ap and C =∅, ∀k ≥r+1 [12,Lemma1.8.1], [3].
1 2 v 2 2 3 2 3 k
3. Amaximalrepresentationofs∈S isanyexpressions= v a n ,a ∈IN,with v a =ord(s)
j=1 j j i j=1 j
P P
andinthiscasewedefinesupport ofsas Supp(s):={n ∈Ap |a 6=0}. Recall:
i 1 i
Supp(s)dependsonthechoiceofamaximalrepresentationofs. Furtherfors∈S,wedefine:
|Supp(s)|:=max |Supp( a n )|, suchthats= a n isamaximalrepresentationofs .
i i i i i i
(cid:8) P P (cid:9)
4. ForasubsetH ⊆S, Supp(H):= Supp(s ),s ∈H .
i i
S(cid:8) (cid:9)
2
5. Wecallinducedbys= v a n (maximalrepresentation)anelement
j=1 j j
s′ = v b n ,with P0≤b ≤a . Recall: ord(s′)= b by[12,Lemma1.11].
j=0 j j j j j
P P
Thefollowingtwopropositionsarecrucialinthesequel.
Proposition1.4 LetS beasinSetting1.2andlets∈C , k ≥2. Then
k
1. Foreachs= a n ∈C (maximalrepresentation,with a =k),wehave:
j≥1 j j k j≥1 j
P P
(a) Everyelements′ = b n inducedbys,with b =h, belongstoC .
j≥1 j j j≥1 j h
(b) Supp(C )⊆Supp(PC )⊆...⊆Supp(C ). P
k k−1 2
2. Ifs=g+ewithg ∈D , (h≤k−1),anymaximalrepresentations= a n hasa =0.
h j≥1 j j 1
P
Proof.(1.a).See[12,Lemma1.11].
(2).Ifa >0,theng =(a −1)e+ a n hasord=k−1,by(1.a),contradiction. ⋄
1 1 j≥2 j j
P
Proposition1.5 LetS beasinSetting1.2,letk beasin(1.3.1)andlet k ≥2.
0
1. H (k)−H (k−1)=|C |−|D | [12,Proposition1.9.3],[3,Remark4.1]
R R k k
2. GisCohenMacaulay ⇐⇒D =∅ foreachk ≥2. [12,Theorem1.6].
k
3. If|D |≤k+1foreveryk≥2,thenH isnon-decreasing [5,Theorem3.3,Corollary3.4].
k R
4. If|D |≥k+1,then |C |≥h+1 forall h∈[2,k] [5,proofofProposition3.9].
k h
5. Inparticular(recall: k =min{k|D 6=∅})[5, Corollary3.11]:
0 k
(a) IfH isdecreasingatlevelk,then|D |≥max{1+|C |, k+2},
R k k
(b) IfH isdecreasing,then|Ap |≥k +1,
R k0 0
(c) IfH isdecreasing,then|Ap |≥3.
R 2
WeshowasemigroupS withH decreasing,whichisthefirstexampleinthesensethate =v+3 =13, |Ap | =3
R 2
areminimalinordertohavedecreasingHilbertfunction,see(4.2),(4.3).
Example1.6 [14,Section2] Let S =<13,19,24,44,49,54,55,59,60,66>,withe=13, v =10,
Ap={ 0 19, 24, 38, 43, 44, 48, 49, 54, 55, 59, 60, 66 }.
M \2M 13 19 24 44 49 54 55 59 60 66 dim = 10
2M \3M 26 32 37 38 43 48 68 73 79 dim = 9
3M \4M 39 45 50 51 56 57 61 62 67 72 92 dim = 11
4M \5M 52 58 63 64 69 70 74 75 80 81 85 86 dim = 12
5M \6M 65 71 76 77 82 83 87 88 93 94 98 99 105 dim = 13
ThenHR =[1,10,9,11,12,13→], HR′ =[1,9,3].FurtherSupp(D2+e)=SuppC2 ={19,24}:
D2 = {44,49,54,59} C2 ={38,43,48}={19·2,19+24,24·2}=Ap2
D +e = {57,62,67,72} 57=3·19, 62=2·19+24, 67=19+2·24, 72=3·24
2
D = {68,73} C ={57,62,67,72}=D +e
3 3 2
D +e = {81,86} 81=3·19+24, 86=2·19+2·24
3
D = {92} C ={81,86}=D +e
4 4 3
D +e = {105} 105=3·19+2·24
4
D = {∅} C ={105}=D +e.
5 5 4
3
2 Technical analysis of C and D via supports and Ape´ry subsets.
k k
Lemma2.1 Letx= v a n ∈C ,witha ≥1,foreachi, v a =ord(x) =kand|Supp(x)|=q.
i=2 i i k i i=2 i
P P
(a) |C |≥hp+1≥q, if q ≥h+1, p≥1
Let 2≤h<k, then h
(cid:20) (b) |Ch|≥q, if q ≤h.
Proof.Firstrecallthat,by(1.4.1a),everyelementoforderhinducedbythegivenmaximalrepresentationofx,belongs
toC . WedenoteforsimplicitySupp(x)={m <m <··· .<m },distinctminimalgenerators,withm 6=eby
h 1 2 q i
(1.4.2).
(a). Ifq ≥h+1.
Thenwecanconstructthe(h+1)+h(q−h−1)distinctinducedelementsinC :
h
i=1,··· ,h+1, for η =1,
σ =( η+hm )−m , η =1,··· ,q−h, .
(cid:26) η,i j=η j i (cid:20) i=η,··· ,h+η−1, for 2≤η ≤q−h, (cid:27)
P
(b). If1≤q ≤h,thereexists(a′,··· ,a′)suchthata ≥a′ ≥1, ∀i=1,··· ,q, a′ =h+1. ThenC contains
1 q i i i i h
theqdistinctelements σj = q1a′imi−mj, j =1,··· ,q. ⋄ P
P
Proposition2.2 Letk =min{k∈N|D 6=∅}, d=max{ord(σ), σ ∈Ap}.
0 k
1. Letg ∈D , g+e= a n ,with a =k+p, p≥1(maximalrepresentation):
k j j j j j
P P
(a) Lety ∈C , h<k+p,beinducedbyg+e;if h≤max{p+1,k }, then y belongstoAp.
h 0
(b) p≤d−1.
(c) |Supp(g+e)|≤|Ap |.
p+1
2. If Ap =∅, then D +e=C foreachk ≥2.
3 k k+1
Proof.1.(a) Let h≤k andletz ∈C \Ap; thenz−e∈S, with ord(z−e)≤h−2,hencez−e∈D , r <k ,
0 h r 0
impossible. Further,ify = b n andy ∈/ Ap,with h= b ≤p+1,theny =e+σ,with 0<σ ∈S. Then:
j j j
g =σ+ (aj −bj)nj =⇒oPrd(g)≥1+k+p−(p+1)=Pk, contradiction.
P
1.(b) Letg+e = a n (maximalrepresentation),if a ≥ k+d,thend+1 ≤ a −k+1andsoevery
j j j j j j j
inducedelement∈PCd+1belongstoApby(1),impossibleP,since(d+1)M ⊆M +e;hePnceord(g+e)=k+1.
1.(c)Let|Supp(g+e)|=qandleth :=p+1< p+k,by(2.1)thereareatleastqelementsinC . Theseelements
h
areinAp,by1.(a).Henceq ≤|Ap |.
p+1
2. Ifthereexistsg ∈D suchthatord(g+e)≥k+2,thenp≥2;by1.(a)wewouldhavey ∈Ap foreveryy ∈C
k 3 3
inducedbyg+e. ⋄
Remark2.3 1. (2.2.1a)cannotbeimproved,forexamplein(1.6)ifs = 92 : ord(s) = 3, 92+e = 105 =
3·19+2·24=⇒92∈D , p=1 and2·19+24=62=49+e∈C \Ap (here3=p+2).
4 3 3
2. Letx ,x ∈D , x 6=x suchthatSupp(x +e)=Supp(x +e), x +e= α n ,
1 2 k 1 2 1 2 1 i i
x2+e= βini. Thenthereexist i,j suchthatαi >βi, αj <βj. P
Proof. IfxP6= x andα ≥ β foreachi,thenx +e = x +e+σ,ord(σ) ≥ 1,henceord(x ) > ord(x ),
1 2 i i 1 2 1 2
impossible.
3. Letx ∈ D , ord(x+e) ≥ k+2. Ify ∈ D andord(y+e) = h ≤ k+1. Theny+ecannotbeinducedby
k k
x+e.
Proof.Ifx+e=y+e+s,withord(s)≥1thenx=y+s,ord(x) >ord(y),impossible,sincebyassumption
ord(x)=ord(y)=k−1.
GiventhesetsC ,C ,withh < k,weestimatelowerboundsforthecardinalityofC ,byenumeratingtheelements
h k h
inducedbyC . Wefirstconsidertheelementsinducedbythesubset{x∈C suchthat|Supp(x)|≤2}.
k k
Lemma2.4 Forx ∈C ,(k ≥3), let x =a n +b n , a +b =kandlet r ∈[2,k−1].
r k r r i r j r r
4
1. Letx ∈C ;thenumberofdistinctelementsofC inducedbyx is
1 k h 1
1+min{b ,h}, if a ≥h
1 1
β1 =1+min{a1,b1,h,k−h}= 1+a1 if a1 <h≤b1
1+k−h if a ,b <h
1 1

withβ ≥1 andβ =1⇐⇒ a b =0.
1 1 1 1
2. Letx ,x bedistinctelementsinC suchthat Supp(x )∪Supp(x )={n ,n }.
1 2 k 1 2 i j
x =a n +b n ,
1 1 i 1 j
Wecanassumethat x2 =a2ni+b2nj,
 a +b =k, 0≤a <a , =⇒ 0≤b <b .
i i 1 2 2 1
ThenumberofdistinctelementsofC inducedby{x ,x }is
h 1 2
(a) 1+min{b ,h}≥2 (=2⇐⇒b =1,b =0) if h≤a <a
1 1 2 1 2
 (b) 2+min{a ,k−h}+min{h−a −1,b }≥3 if 0≤a <h≤a
β = 1 1 2 1 2
2
(c) 2+min{a ,k−h}+min{a −a −1,k−h}≥2 if 0≤a <a <h
 1 2 1 1 2
 β =2⇐⇒a =0,a =1
 2 1 2
x =a n +b n
1 1 i 1 j
3. If C ⊇{x ,x ,x }, let x2 =a2ni+b2nj, ,
k 1 2 3  x3 =a3ni+b3nj,
a +b =k, 0≤a <a <a , b >b >b ≥0
i i 1 2 3 1 2 3
ThenumberofdistinctelementsofC inducedby{x ,x ,x }is
h 1 2 3
(a) 1+min{b ,h}≥3 if h≤a <a <a
1 1 2 3
(b) 2+min{a ,k−h}+min{h−a −1,b}≥3 if a <h≤a <a
1 1 2 1 2 3
 further β ≥4, except the cases:
 3
 (i) h=2, a ∈{0,1}
 1
 (ii) k =h+1, a =h−1, b =2
 1 1
 (iii) a =0, b =1
 1 2
β =
3 (c) 3+min{a ,k−h}+min{a −a −1,k−h}+min{h−a −1,b } if a <a <h≤a
 1 2 1 2 3 1 2 3

 further β ≥4, except the cases:
 3
 (c ) a =a −1=min{h−2,b }=0
 1 1 2 3

(d) 3+min{a ,k−h}+ min{a −a −1,k−h} if a <a <a <h
 1 i=1,2 i+1 i 1 2 3

 further β =3⇐⇒aP=0,a =1, a =2
 3 1 2 3

Proof.1. Puta =a,b =bforsimplicity.Ifa≥h,theinduceddistinctelementsinC are:
1 1 h
{hn , (h−1)n +n ,··· ,hn },ifh≤b,
i i j j
{hn , (h−1)n +n ,··· ,bn },ifh>b.
i i j j
Thenβ =1+min{b,h}=i+min{a,b,h,a+b−h}.
1
If a < h ≤ b, then {hn , (h−1)n +n ,··· ,(h−a)n +an } are indueced distinct elements of C . Hence
i i j i j h
β =a+1wherea=min{a,b,h,a+b−h}sincea≤a=b−h<b.
1
Ifa,b<h, thentheinduceddistinctelementsare:an +(h−a)n , (a−1)n +(h−a+1)n ,··· ,(h−b)n +bn ,
i j i j i j
then β = a+b−h+1 = k−h+1, furtherβ = 1+min{a,b,h,a+b−h} = a+b−h(= k−h), since
1 1
a+b−h<a<h, a+b−h<b.
If a,b ≥ h, then β = 1+h and the induced distinct elements are {hn , (h−1)n +n ,··· ,hn }. In this case
1 i i j j
|C |=h+1ismaximal. Then,ifa≥h,β =1+min{b ,h}.
h 1 1
2(a). Theinduceddistinctelements∈C are {(h−i)n +in , 0≤i≤min{b ,h} }.
h i j 1
2(b). Wehave1+min{a ,k−h}elements∈C inducedbyx , by(1):
1 h 1
(a −i)n +(h−a +i)n , 0≤i≤min{a ,k−h}
1 i 1 j 1
Moreover fromx wecanextractthefollowingdistinctadditionalelements
2
5
(h−p)n +pn , p≥0, h−p≥a +1, p≤b hencewith 0≤p≤min{h−a −1,b }.
i j 1 2 1 2
(cid:0) (cid:1)
2(c). First, x induces the same elements of C considered in (2.b); moreover from x one gets other M distinct
1 h 2
elements
(a −p)n +(h−a +p)n ,withp≥0, a −p≥a +1, h−a +p≤b
2 i 2 j 2 1 2 2
(hence0≤p≤min{a −a −1,k−h}),where
2 1
eitherM =(a −a ), if h−a −1≤b (elements(a −p)n +(h−a +p)n ,with0≤p≤a −a −1);
2 1 1 2 2 i 2 j 2 1
or M =k−h+1, if h−a −1>b (elements a n +(h−a )n ,··· ,(h−b )n +b n ).
1 2 2 i 2 j 2 i 2 j
3. Itcomesdirectlybyusingthesameideasoftheproofofstatement(2). ⋄
Thislemmaallowstoprovethemoregeneral
Proposition2.5 Assumek ≥3, 2≤h≤k−1and C ⊇{x ,x ,··· ,x }, p≤k+1, with
k 1 2 p
x =a n +b n , 1≤i≤p, a +b =k
i i i i j i i
(cid:26) 0≤a1 <a2 <···<ap ≤k (=⇒k ≥b1 >b2 >···>bp ≥0)
and let β be the number of distinct elements of C induced by {x , i = 1,··· ,p}. Then β ≥ min{h+1,p},
p h i p
precisely:
1+min{b ,h} if h≤a <···<a
1 1 p
 i+1+min{a1,k−h}+···+min{ai−ai−1−1,k−h}+min{h−ai−1,bi+1} if <ai <h≤ai+1 <
(i≤p−1)

 p+min{a1,k−h}+···+min{ap−ap−1−1,k−h} if <ap <h

Lemma2.6 Letx ,x bedistinctelementsinD suchthatSupp(x +e)={n ,n }, Supp(x +e)={n ,n }, {n ,n }∩
1 2 k 1 i j 2 t u i j
y =x +e=an +bn
1 1 i j
{nt,nu}=∅. Let y2 =x2+e=cnt+dnu .
 abcd6=0, a+b=k+r , c+d=k+r
1 2
Let r := min{r ,r }, and let 2 ≤ h ≤ k + r. Consider the induced elements z = pn + (h − p)n , z =
1 2 1 i j 2
qn +(h−q)n ∈C .
t u h
Then z 6= z for every p,q,h and |C | ≥ β + β where β ,β are defined in (2.4.1) (called β ).
1 2 h ab cd ab cd 1
Consequently
|C |≥4, if h<k+r, |C |≥3, if h=k+r, r 6=r , |C |≥2, if h=k+r, r =r .
h h 1 2 h 1 2
Proof.Assumez =z ,thenbysubstitutingweget
1 2
y =(a−p)n +(b+p−h)n +z =(a−p)n +(b+p−h)n +qn +(h−q)n
1 i j 1 i j t u .
(cid:20) y2 =(c−q)nt+(d+q−h)nu+z1 =pni+(h−p)nj +(c−q)nt+(d+q−h)nu
Firstnotethatbytheassumption,wehavez 6=z if: p=q =0,orp=q−h=0,orq =h−p=0,orh=p=q;
1 2
moreoverifthreecoefficientsare6=0,then|Supp(y )|=3,againsttheassumption.Thisargumentallowstocomplete
i
theproofintheremainingcaseswhicharethefollowing:
(a−p)n (b+p−h)n qn (h−q)n | pn (h−p)n (c−q)n (d+q−h)n
i j t u i j t u
 6=0 0 0 6=0 | 6=0 b c d−h
6=0 0 6=0 0 | 6=0 b c−h d

 0 6=0 0 6=0 | 6=0 6=0 c d−h

 0 6=0 6=0 0 | 6=0 6=0 c−h d ⋄

Lemma2.7 Letk≥3andletx ,x bedistinctelementsinD suchthat|Supp(x +e)|=|Supp(x +e)|=2,
1 2 k 1 2
x +e=an +bn
1 i j , a,b,c,d>0with n ,n ,n distinctelementsinAp .
(cid:26) x2+e=cnt+dnj i j t 1
Forh<k,considertheinducedelementsinC : z =pn +(h−p)n , z′ =qn +(h−q)n . Then
h i j t j
(cid:8) (cid:9)
(i) pq =0 and p+q >0
 (ii) a<h and q ≥max{1,h−d}
1. z 6=z′inthefollowingcases
or

 c<h and p≥max{1,h−b}

6
x +e=n +bn
2. Inthecases 1 i j wehave|C |≥3, intheremainingcases|C |≥4.
(cid:26) x2+e=nt+dnj h h
Proof.1. (i)isimmediatebytheassumptions.
(ii). Itisenoughtoconsiderp>0by(i);ifz =z′,then
x +e=pn +(d+q−p)n +(c−q)n ,whered+q−p≥h−p≥1,sincep≤a<h. Then:
2 i j t
|Supp(x +e)|≥3 if c−q >0
2 , contradictioninanycase.
(cid:20) Supp(x2+e)=Supp(x1+e) if c−q =0
2. Wecanassume0<a≤c. Thefollowingz belongtoC andaredistinct.
i h
(ii) 1≤a,c<h: z =an +(h−a)n
(i) a=c=1: z =n +(h−1)n 1 i j
1 i j (ac≥2) z =(a−1)n +(h−a+1)n
z =h n 2 i j
2 j z =cn +(h−c)n
z =n +(h−1)n 3 t j
3 t j z =(c−1)n +(h−c+1)n
4 t j
(iii) a<h=c: z =an +(h−a)n
1 i j
(iv) a<h<c: z =an +(h−a)n
z2 =(a−1)ni+(h−a+1)nj 1 i j
z =(a−1)n +(h−a+1)n
z3 =c nt 2 i j
z =h n
z4 =(c−1)nt+nj 3 t
z =(h−1)n +n
4 t j
(v) h≤a≤c: z =hn
1 i
z =(h−1)n +n
2 i j
z =h n
3 t
z =(h−1)n +n
4 t j
Thenontrivialsubcasesof(i),··· ,(iv)comedirectlyfrompart1.
Case(v).z =z =⇒ eitherSupp(x +e)=Supp(x +e)(ifa=h),or|Supp(x +e)| =3,(ifa >h),against
1 4 1 2 1
theassumptions.
z = z =⇒ either Supp(x +e) = Supp(x +e)( if c = h), or |Supp(x +e)| = 3,( if c > h), against the
2 3 2 1 2
assumptions. ⋄
Nextpropositionshowsthatfork=2,statement(1.5.4)holdsalsoatstepk+1.
Proposition2.8 AssumeH decreasingatlevel h. Then
R
1. Thereexistx ,x ∈D suchthat|Supp(x +e)|≥2, for r =1,2.
1 2 h r
2. |C |≥4.
3
Proof. 1. AssumeH decreasingatlevelh: weknowthatm =|C |≤|D |−1. If|Supp(x +e)| = 1 ∀x ∈D ,
R h h r r h
thenx +e=α m , α ≥h+1,m ∈Ap ,withm 6=m , if r6=s. Hencetheelements{hm }wouldbedistinct
r r r r r 1 r s r
elements∈C and|C |>m.
h h
Henceletx ∈D ,|Supp(x +e)|≥2,withamaximalrepresentation x +e= β m ,withβ ≥ 1,m ∈Ap
1 h 1 1 i i i i 1
distinctelements.Oneinducedelement∈Chis c1 =mi+mj+sforsomesofordPerh−2. Ifeachxi ∈Dh,i>1
has maximalrepresentationof the type x +e = α n , then hm ,··· ,hm ∈ C are distinct elements; further
i i i j m+1 h
hm 6= c for each i (otherwise x +e = c +(α −h)m and so |Supp(x +e)| ≥ 2). Then |C | ≥ m+1,
i 1 i 1 i i i h
contradiction.
2. By(1.5.3and4)weknowthatifh > 2then|C |≥ j+1forall2 ≤ j ≤ h. Itremainstoprovethestatementfor
j
h=2, and |D |≥4by[5,prop.2.4].
2
SinceC =Ap ∪(D3+e),thefactistrueiford(x+e)=3 ∀x∈D . Henceweassumethereexistsx∈D such
3 3 2 2 2
thatord(x+e)≥4.
Inthisproof,forsimplicity,wedenotetheelementsofSupp(x)withm ,m ,m ··· andassumem <m <m .
1 2 3 1 2 3
If|Supp(x+e)|≥4:x+e=m +m +m +m +s,withm <m <m <m ,then|C |≥4by(2.1.a).
1 2 3 4 1 2 3 4 3
If|Supp(x+e)|=3, x+e=β m +β m +β m withβ ≥1, β ≥5),wefind4inducedelementsin C as
1 1 2 2 3 3 i i 3
follows: P
whenβ ≥2,β ≥2:{m +m +m ,2m +m ,2m +m ,2m +m };
1 2 1 2 3 1 2 1 3 2 3
7
whenβ ≥2,β ≥2:{m +m +m ,2m +m ,2m +m ,m +2m };
1 3 1 2 3 1 2 1 3 2 3
whenβ ,β ≥2,β ≥2:{m +m +m ,2m +m ,m +2m ,m +2m };
1 2 3 1 2 3 2 3 1 2 2 3
whenβ ≥3:{m +m +m ,2m +m ,2m +m ,3m }.
1 1 2 3 1 2 1 3 1
If|Supp(x+e)|=3, x+e=β m +β m +β m withβ ≥1, β =4,thenx+einducesinC threedistinct
1 1 2 2 3 3 i i 3
elementsc1,c2,c3(seethefollowingtable).TogetthefourthelemenPt,wetakey ∈D2,y 6=x,|Supp(y+e)|≥2(by
1). IfSupp(x+e) 6= Supp(y+e)andeachelementinducedbyy+einC isalsoinducedbyx+e,then wecan
3
alwaysreducetothecaseSupp(x+e)⊇Supp(y+e),bysuitablesubstitutions(inoneormoresteps).
Hence we can assume Supp(x+e) ⊇ Supp(y +e); we study in the next table one of the possible subcases, the
remainingaresimilar. Let
x+e=m +2m +m
1 2 3
(cid:26) y+e=β1′m1+β2′m2+β3′m3, βi′ ≥3
If β′ =3,theny+e6=zforeachz ∈C inducedbyx+e,otherwisePord(x)>ord(y). Henceassume β′ >3;
i 3 i
P c1 =m1+2m2 P
 c2 =m1+m2+m3
c =2m +m
3 2 3

 and, according to the values ofthe β′,
 i
 c =2m +m if β′ =2,β′ ≥1
thenC ⊇ 4 1 2 1 2
3  c5 =m2+2m3 if β1′ ≤1,1≤β2′ ≤2,β3′ =2
 c =3m if β′ ≥3
 6 3 3
 c7 =3m1 if β1′ ≥3
 c8 =3m2, or 2m1+m2, or m2+2m3 if β2′ ≥3
 c =2m +m if β′ =2,β′ =0,β′ ≥1
 9 1 3 1 2 3
Clearlytheelementsc , i = 4,··· ,7aredistinctfromc ,c ,c . Incaseβ′ ≥ 3wehaveeithery+e = 3m +m ,
i 1 2 3 2 2 i
(i = 1 or3) or y+e = 3m . Inthe first case if 3m = m +m +m , then2m = m +m =⇒y+e =
2 2 1 2 3 2 1 3
2m +m +m , or y+e=m +m +2m andwecanaddc resp. c .
1 2 3 1 2 3 4 5
In case y +e = 3m , we get3m ∈/ {c ,c ,c }, otherwise 2m = m +m =⇒y +e = m +m +m , i.e.,
2 2 1 2 3 2 1 3 1 2 3
x=y+m ,impossiblesinceord(x) =ord(y).
2
Ifβ′ =2,β′ =0,β′ ≥1,thenc ∈/ {c ,c ,c }otherwise2m +m =m +2m =⇒m +m =2m =⇒x−y =
1 2 3 9 1 2 3 1 3 1 2 1 3 2
(2−α)m ,impossibleinanycase.
3
Finallyweassume: |Supp(x+e)| = |Supp(y+e)| = 2, ord(x+e) ≥ 4. Bylemmas(2.6)and(2.7),itremainsto
analysethecases
x+e=m +bm x +e=am +bm
1 2 1 1 2 x+e=am +bm a+b=3
(I) by≥+e3,=dm≥32+dm2 (II) xa2++be≥=4,cmc+1+dd≥m32 (III)(cid:26) y+e=cm31+dm42 c+d≥4
(I). Theinduceddistinctelementsarec1 =m1+2m2 c2 =3m2 c3 =2m2+m3 .
Byassumption,thereexisttwootherelem(cid:8)entsz1,z2 ∈D2;bytheabovetoolsandby(2.3),w(cid:9)ecanrestricttothecases
(i) z +e=αm +βm α6=1 α+β ≥3
1 1 2
b≥3, d≥2: (ii) z1+e=αm3+βm2 α6=1 α+β ≥3 .
(iii) z +e=αm +βm α+β ≥3
1 1 3

In the first two cases, if α > 0, then α ≥ 2 (by 2.3.2), thereforeif β > 0 we obtain |C | ≥ 4 by applying(2.7.2)
3
to thepair of elements{z +e, y +e} (resp{x+e, z +e}). Incase (iii), again, from(2.7.2), bysubstitutingin
1 1
{x+e, z +e}m′ =m ,m′ =m ,m′ =m ,wededuce|C |≥4, if αβ >0. Theremainingpossibilitiesare
1 1 2 2 1 3 3 3
(i′) z +e=βm β ≥3
1 2
(ii′) z +e=αm α≥3 .
1 1
(iii′) z +e=αm α≥4
1 3
Incase(i′)wecanassumetheelementz ∈D ,verifiesz +e=αm , α≥3 (orz +e=αm ).
2 2 2 1 2 3
If3m = 2m +m , then, either|Supp(z +e)| = 3 andwe aredone, or z +e = 2m +m , impossiblesince
1 2 3 2 2 2 3
y+e=m +dm .
3 2
Similarlywecansolvecase(ii′).
Incase(iii′),if3m =m +2m , then z +e=m +2m +(α−3)m .Thecaseα≥5hasalreadybeenproved
3 1 2 1 1 2 3
above.Ifα=4,weconsiderm +m +m ∈C which,inthissituation,isdistinctfromc ,c ,c .
1 2 3 3 1 2 3
8
(II). If we have 4 elementswith support ⊆ {m ,m }, then we are doneby (2.5). The othercases can be studied
1 2
amongtheonesofshape(I)or(III).
(III). By 2.6, it remains to consider elements z + e = αm + βm , with α + β = 3. In this case, C ⊇
i 1 2 3
{x+e,z +e,}∪{y ,y , inducedbyy+e}. ⋄
1 1 2
3 Structure of C .
3
Theorem3.1 Assume |Ap |=3. Then
2
1. If|C |≥2, then|Supp(x)|≤2, foreachx∈C .
3 3
2. If|C |≥4, then
3
(a) Thereexistx ,x ∈C suchthat|Supp(x )|=2.
1 2 3 i
3n
2n i
(b) Thereexistni,nj ∈Ap1suchthatC2(=Ap2)= n2nii+nj, C3 = 2nnii++2nnjj .
j 3n
j
 
Proof.1. Ifx∈C ,then|Supp(x)|≤3. Firstweshowthatifx ,x ∈C ,then|Supp(x )|≤2fori=1,2.Assume
3 1 2 3 i
thatx =n +n +n , n <n <n . Then,by(1.4.1a)andtheassumption,wededucethat
1 i j k i j k
Ap ={n +n ,n +n ,n +n } (∗)
2 i j i k j k
Hence2n ,2n ∈/ Ap, 2n ∈Ap⇐⇒n +n =2n . Ifx ∈C ,x 6=x therearethreecases:
i k j i k j 2 3 2 1
n +n
i j
(a) If |Supp(x2)|=1,hencex2 =3nt,then2nt ∈C2,by(1.4.1a).Henceby(∗)2nt = ni+nk
n +n
j k

Ineverycaseweseethat|Supp(x )|≥2,acontradiction.
2
(b) If |Supp(x )|=2,x =2n +n ,t6=v,then2n ∈C ,by(1.4.1a).
2 2 t v t 2
n +n =⇒x =n +n +n =⇒n ∈{n ,n }
i j 2 i j v v i j
 nv =ni=⇒x2 =2ni+nj =⇒2ni ∈Ap2
Henceby(∗)2n = n =n =⇒x =n +2n =⇒n +n =2n ∈Ap =⇒2n ∈Ap
t  v j 2 i j i k j 2 i 2
 ni+nk (similar)

 nj +nk (similar)

everycasecontradictsequality(∗).
(c) If |Supp(x )| = 3, x = n +n +n , n < n < n , by (∗)wemusthave: n +n = n +n ,hence
2 2 t v w t v w t v i j
x =n +n +n . Thisequalityimpliesthat
2 i j w
n 6=n (sincex 6=x )
w k 2 1
 nw 6=nj (otherwise x2 =ni+2nj =2ni+nk, see(∗)) .
 n +n ∈Ap =⇒n +n =n +n
i w 2 i w j k
Hencex =2n +n =n +2n ,contradiction.
2 j k i k
2(a). Let x ∈ C ,1 ≤ i ≤ 4, distinct elements such that 1 ≤ |Supp(x )| ≤ 2 (by 1). If |Supp(x )| = 1, for
i 3 i i
x = 3n
1 a
i=1,2,3, x2 = 3ni ,thenby(1.4.1a)C2 =Ap2 ={2na,2ni,2nc}. Since|Ap2|=3,by(1)and(1.4.1a),we
 x = 3n
3 c
have|Supp(x )|=2:wecanassumex =2n +n ,n +n =2n . Hencex =n +n +n ,withn 6=n and
4 4 a d a d i 2 a d i a d
so|Supp(x )|≥2,againsttheassumption.
2
x =2n +n , n 6=n
2(b). Let{x ,x ,x ,x }⊆C ;by2(a),wecanassumethat 1 i j i j .
1 2 3 4 3 (cid:26) x2 =2nh+nk nh 6=nk
Weprovethat|Ap |=3=⇒n =n ,n =n , i.e. Supp(x )=Supp(x ),x =n +2n .
2 i k j h 1 2 2 i j
Notethat|Ap | = 3=⇒|{n ,n ,n ,n }| ≤ 3,otherwise,{2n ,2n ,n +n ,n +n } ⊆ Ap =⇒|Ap | ≥ 4. In
2 i j h k i h i j h k 2 2
factif2n = n +n =⇒|Supp(x +e)| ≥ 3(theothercasesaresimilar). Hencetherearethreepossibledistinct
i h k 1
9
situations:
x =2n +n , n 6=n ,
1 i j i j x =2n +n , n 6=n x =2n +n , n 6=n
(b1) x =2n +n nni 6=6=nnh , (b2)(cid:26) x21 =2nhi +nji nhj 6=nii , (b3)(cid:26) x12 =2nii+njk nik 6=njj .
2 h j h j
(b ).WehaveAp ⊇{2n ,n +n ,2n ,n +n },hence|Ap |=3=⇒
1 2 i i j h h j 2
either n +n =2n =⇒x =n +2n (thesis)
i j h 2 i j
(cid:20) or 2ni =nh+nj (analogous).
(b ). Wehave Ap ⊇{2n ,n +n ,n +n ,2n }=⇒2n =n +n =⇒ eitherx =x
2 2 i i j h i h h i j 2 1
(againsttheassumption),or n =n (=⇒x =2n +n ).
h j 2 j i
(b ). Thiscasecannothappen.InfactwehaveAp ={2n ,n +n ,n +n }. Bysimilarargumentsasaboveonecan
3 2 i i j i k
seethatforanyotherelementx∈C , x6=x ,x ,themaximalrepresentationofxmustbewrittenasx=3n . Infact
3 1 2 i
theotherpossiblerepresentationsareincompatible;forinstance,x=3n ,with2n =n +n =⇒x=n +n +n ,
j j k i i j k
impossibleby(2).Thiswouldmeanthat|C |≤3,againsttheassumption.
3
AccordingtotheabovefactswededucethatC ={3n ,2n +n ,n +2n ,3n }. ⋄
3 i i j i j j
Example3.2 Accordingtothenotationof(3.1),weshowseveralexamplesofsemigroupswhichverifytheassump-
tionsof (3.1): Ap = {2n ,n +n ,2n }. The first exampleshowsthatthe conditions(3.1.2b)are, in general, not
2 i i j j
sufficienttohaveH decreasing.
R
1. Let S =< 19,21,24,47,49,50,51,52,53,54,55,56,58,60 > and let n = 21,n = 24. Then: Ap =
i j 2
{42,45,48}, Ap ={63},Ap ={84}, v =e−5, H =[1,14,14,14,16,18,19→]isnon-decreasing.
3 4 R
2. LetS =<19,21,24,65,68,70,71,73,74,75,77,79>withn =21,n =24. Then:
i j
ℓ = 2,Ap = {42,45,48}, Ap = {3n ,2n +n ,n +2n ,3n }, v = e−7, D +e = {4n ,3n +
2 3 i i j i j j 2 i i
n ,...,n +3n }, D ={65,68,71,74,77} H decreasesatlevel2, H =[1,12,10,11,15,18,19→].
j i j 2 R R
3. LetS =<19,21,24,46,49,51,52,54,55,56,58,60>withn =21,n =24. Then:
i j
ℓ=3, |Ap |=4, v =e−7, H decreasingatlevel3,H =[1,12,15,14,16,18,19→].
3 R R
4. S =<30,33,37,64,68,71,73,75,76,78,79,80,82→89,91,92,95>withn =33,n =37. Then:
i j
Ap = {2n ,n +n ,2n }, Ap = {3n ,n +2n ,3n }, Ap = {132= 4n }, v = e−7, H decreasesat
2 i i j j 3 i i j j 4 i R
level4:H =[1,23,25,25,24,27,28,29,30→].
R
Lemma3.3 Let d = max{ord(σ)| σ ∈ Ap}. Assume there exists 3 ≤ r ≤ d such that |Ap | = 1. Let r :=
r 0
min{j | |Ap |=1} (r ≥3), then:
j 0
1. Ifr <d,thereexistsn ∈Ap suchthatAp ={kn } for r ≤k ≤d(and kn ∈Ap , fork <r ).
0 i 1 k i 0 i k 0
2. Ifthereexistsg ∈D , (k ≥ 2)suchthatord(g+e) =k+p, withp ≥ r −1,then Ap = dn (n ∈Ap ),
k 0 d i i 1
and g+e=(k+p)n >dn ; furthersuchelement g isunique.
i i
Proof.1.BytheAdmissibilityTheoremofMacaulayforHR′,weknowthat|Apk|=1,forr0 ≤k ≤d. Nowsuppose
r < d, and|Supp(σ)| ≥ 2, σ ∈ Ap . Ifn ,n ∈ Supp(σ),then,by(1.4),theelementsσ−n , σ−n wouldbe
0 d i j i j
distinctelementsinAp , contradiction.
d−1
2. By the assumption and (2.2.1c), we have r ≤ p + 1 ≤ d and |Supp(g + e)| ≤ |Ap | = 1. Therefore
0 p+1
g+e = (k +p)n , with n ∈ Ap . Then forr ≤ j ≤ p+1, the inducedelementjn ∈ C belongsto Ap , by
i i 1 0 i j j
(2.2.1a). ThenAp = {jn } forr ≤ j ≤ d. Hencek+p > d. Foreach k, if such g exists, then it is uniqueby
j i 0
(2.3.2). ⋄
Proposition3.4 Assume|Ap | = 3,|Ap | = 1andH decreasing. Letℓ = min{h|H decreasesatlevelh}and
2 3 R R
letd=max{ord(σ)| σ ∈Ap}. Wehave:
10