Dimension, depth and zero-divisors of the algebra of basic $k$-covers of a graph

0.25 MB·English

by  Bruno Benedetti

Dimension, depth and zero-divisors
of the algebra of basic k-covers of a graph.
Bruno Benedetti Alex Constantinescu‡‡ Matteo Varbaro§
∗
Inst.Mathematics,MA6-2 DipartimentodiMatematica DipartimentodiMatematica
9
0 TUBerlin Universita´diGenova Universita´diGenova
0
D-10623Berlin,Germany I-16146Genoa,Italy I-16146Genoa,Italy
2

n [email protected] [email protected] [email protected]
a
J
January25,2009

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2

]
C
Abstract
A
. Westudythebasick-coversofabipartitegraphG;thealgebraA¯(G)theyspan,firststudied
h
t byHerzog,isthefiberconeoftheAlexanderdualoftheedgeideal. Wecharacterize when
a
m A¯(G) is a domain in terms of the combinatorics of G; if follows from a result of Hochster
that when A¯(G) is a domain, it is also Cohen-Macaulay. We then study the dimension of
[
A¯(G) by introducing a geometric invariant of bipartite graphs, the “graphical dimension”.
1
WeshowthatthegraphicaldimensionofGisnotlargerthandim(A¯(G)),andequalityholds
v
5 in many cases (e.g. when G is a tree, or a cycle). Finally, we discuss applications of this
9
theorytothearithmetical rank.
8
3
.
1
Introduction
0
9
0 To each homogenous ideal I of a standard graded K algebra S, we can associate a second
:
v standardgraded K-algebra, theso-called fiberconeofI,defined as
i
X
r Fm(I):= Ik/mIk,
a
Mk 0
≥
wherem= S isthemaximalirrelevantideal ofS.
k>0 k
⊕
Fiber cones have been introduced by Northcott and Rees [15] and extensively studied in
literatureeversince. Some typical questions(in general hard to solve)are the following: given
an idealI,
⊲ is F (I)adomain? is itCohen-Macaulay?
m
⊲ what is theKrull dimensionof F (I)? (This invariant is knownas the“analytic spread
m
ofI”.)
SupportedbyDFGviatheBerlinMathematicalSchool;partiallysupportedbyPRAGMATIC08.
∗
‡‡SupportedbyPhdofUniversita´diGenova;partiallysupportedbyPRAGMATIC08.
§SupportedbyPhdofUniversita´diGenova;partiallysupportedbyPRAGMATIC08.
1
At the summer school “Pragmatic 2008”, with Herzog and Welker as teachers, we studied
the special case where S is the polynomial ring in n variables, and I is a monomial ideal asso-
ciated to a bipartite graph G (more precisely, I is the Alexander dual of the edge ideal of G).
Wewilldenoteby A¯(G)thefiber coneofthisideal I. Consistentlywiththegeneral picture, we
focusedon thefollowingquestions:
⊲ When isA¯(G)a domain? When is itCohen-Macaulay?
⊲ What isthedimensionofA¯(G)?
Hereis asummaryofsomeoftheresultscontained inthepresentpaper.
We give a simple, purely combinatorial characterization of those bipartite graphs G for
which A¯(G) is a domain (Theorem 1.8). Namely, for every non-isolated vertex i of G there
mustbean edge i, j in G suchthat
{ }
If i,i isan edgeofG and j, j is an edgeofG,then i, j is alsoan edgeofG.
′ ′ ′ ′
{ } { } { }
In case G satisfies this condition, we show via a theorem of Hochster that A¯(G) is actually a
Cohen-Macaulay normal domain (Proposition 1.1). Moreover we prove in Theorem 3.3 that
A¯(G) is an Hibi ring provided that G is unmixed (in this case one can easily see that A¯(G) is a
domain). This has some interesting consequences: for example it allows us to characterize all
Cohen-Macaulay bipartitegraphsforwhich A¯(G) isGorenstein(Corollary3.6).
WhenA¯(G)isnotadomain,itmaybenon-Cohen-Macaulayaswell. However,weshowthat
depth(A¯(G)) 2 for all bipartite graphs G: see Theorem 1.14. This fact has some interesting
consequences≥,amongwhichincludethattheprojectiveschemeProj(A¯(G))isalwaysconnected.
Anyway we show an example of a bipartite graph G for which A¯(G) is not equidimensional,
andthusnot Cohen-Macaulay (Example2.15).
Furthermore: we prove that the dimension of A¯(G) is n when G is a cycle (Proposition
2
2.12), and n +1 when G is a path (Corollary 2.8) or a Cohen Macaulay graph (Corollary
⌊2⌋
2.9). Theseresultsare nottrivial,asthe“obvious”approachtotheproblem– thatis,toasymp-
totically count basic k-covers1 – leads to long, nasty calculations. We followed in fact another
road, introducing a numerical invariant – the “graphical dimension gdim(G)” – for each bipar-
titegraph G.
Foragenericbipartitegraph G,we showthat
gdim(G) dimA¯(G) (Theorem 2.7).
≤
Indeed, fortrees andcycles, onehas equality:
gdim(G)=dimA¯(G) (Theorem 2.22and Proposition2.12).
Wedo notknowwhetherequalityholdstrueingeneral, ornot.
Finally,inthelastsection,wediscuss:
the dimension issue in a more general setting (that considers hypergraphs instead of bi-
•
partitegraphs);
unmixedgraphsand Hibirings;
•
applicationstothearithmeticalrank oftheAlexanderdualoftheedgeidealofG.
•
1Seethenextparagraphforthedefinition.
2
0.1 Definitions and notation
0.1.1 Bipartitegraphs andvertex covers
Let G be a simple graph on n vertices, labeled by 1, ...,n. A vector a=(a ,...,a ) a k-cover
1 n
ofGif:
k and all a’s arenonnegativeintegers;
i
•
notalla ’sare zero;
i
•
foreveryedge i, j ofG onehas a +a k.
i j
• { } ≥
This concept generalizes the graph theoretical notion of vertex cover, which is a subset of
[n] that has non-empty intersection with all the edges of G. Vertex covers in our language are
just1-coverswhoseentriesare either0or1.
Two covers can be “summed” vertex-wise: a k-cover plus a k -cover gives a (k+k )-cover.
′ ′
A k-cover a is said to be decomposable if a = b+c, for a suitable h-cover b and a suitable
(k h)-cover c. A k-cover a is said to be non-basic if it can be decomposed into a k-cover a,
′
−
and a 0-cover a . Indecomposable (resp. basic) is of course contrary to decomposable (resp.
′′
non-basic). We also say that a k-cover a can be “lopped at i” if replacing a with a 1 in the
i i
−
vectora, givesagainak-cover.
Example0.1. LetGbeahexagon,withtheverticeslabelledclockwise. Then,a=(1,0,1,1,0,1)
is a basic 1-cover; a = (1,1,0,1,1,0) is also a basic 1-cover; yet their sum c = a+b =
(2,1,1,2,1,1) is a 2-cover that is not basic (c can be lopped at 1). This shows that A¯(G) is
notadomainwhen Gis ahexagon: a and bare bothzero-divisors.
The algebra has however depth at least 2, since we will see that d = (1,0,1,0,1,0) and
e = (0,1,0,1,0,1) are non-zero-divisors and form a regular sequence. As a consequence of
Theorem2.12we willsee thatthedimensionofA¯(G) is3.
Recall that a graph G is bipartite if its vertex set [n] := 1,...,n can be partitioned into
{ }
subsetsA,BsuchthatalltheedgesofGlinkonevertexinAwithonevertexinB. Incaseevery
vertex in A is adjacent to every vertex of B, we say that G is bipartite complete, and we write
G=K ,wherea= A andb= B . AgraphGisbipartiteifandonlyifGdoesnotcontainan
a,b
| | | |
oddcycle. Moreover,thedisjointunionofk graphsG ,...,G isbipartiteifandonlyifeachG
1 k i
is bipartite. Finally, it was established in [8, Theorem 5.1]) that, providing the graph G has at
leastoneedge:
1. Ifa isak-coverwithk 3, ais decomposable;
≥
2. thegraph G isbipartiteifand onlyifG has no indecomposable2-covers.
Notice that a bipartite graph might admit a basic k-cover even when k 2: for example, if G
≥
consists in two vertices and a single edge, and a is the vector (0,k), clearly a is decomposable
as a=(0,1)+(0,k 1),buta isbasic.
−
0.1.2 The Alexanderdual ofthe edge idealofa graphand itsfiber cone
Let S=K[x ,...,x ] be the polynomial ring on n variables over a field K, and let t be an extra
1 n
indeterminate. Let G beagraph onn vertices. Denoteby
A(G) S[t]
k
⊆
theinfinitedimensionalK-vectorspacegenerated bythe
xa1 xantk such thata isak-coverofG .
{ 1 ··· n }
3
Thegraded S-algebra
A(G)= A(G)
k
Mk 0
≥
iscalled thevertex cover algebraofG.
If G is bipartite and has at least one edge, it admits no indecomposable k-covers for k 2;
≥
thusA(G)isastandard graded S-algebra[8]. In particular, thequotient
A¯(G)=A(G)/mA(G)
isastandard graded K-algebra. It iseasy to check that:
A(G) is generated as a S-algebra by the monomials xa1 xant such that a=(a ,...,a )
• 1 ··· n 1 n
isan indecomposable1-cover;
A¯(G) is generated as a K-algebra by the monomials xa1 xant such that a=(a ,...,a )
• 1 ··· n 1 n
isan indecomposable1-cover;
a basis of A¯ (G) as a K-vector space is given by the monomials xa1 xantk such that
• k 1 ··· n
a=(a ,...,a )is abasick-cover;
1 n
in particular, the Hilbert function HFA¯(G)(k) = dimKA¯(G)k counts the number of basic
•
k-coversofG.
RecallthatforahomogeneousidealI SthedimensionofitsfiberconeF (I)iscalledthe
m
⊆
analytic spread of I. When K is an infinite field, the analytic spread is the minimal number of
generators ofany idealJ satisfying
– J iscontained inI;
– thereexistsapositiveintegers suchthat Js I =Is+1;
·
– J isminimalamong theideals thatsatisfytheprevioustwoproperties.
In particular,theanalyticspread islarger orequalthan thearithmeticalrank, defined as
ara(I):=min numberofgenerators ofJ s.t. √J =√I .
{ }
Now,letG beabipartitegraph withat leastoneedge; theideal
I = x, x ;
i j
i,j ed\geofG(cid:0) (cid:1)
{ }
is a squarefree monomial ideal, whose fiber cone coincides with A¯(G). In other words, A¯(G)
is the fiber cone of the Alexander dual of the edge ideal of G, and dim(A¯(G)) is its analytic
spread. (Also, A(G)correspondstotheRees algebraS[It].)
Notes.
1. If G is not bipartite, it is still possible to define A¯(G) as the symbolic fiber cone of I,
provided G has at least one edge; see the last section for a broader approach involving
hypergraphs.
2. If G is the disjoint union of n points, G is bipartite but A¯(G) is not finitely generated as
K-algebra. The basic k-covers of G are those that assign 1 to a single vertex and 0 to all
othervertices;twicesuchabasiccoveris notabasiccoveranymore.
3. Throughout thispaper wealwaysassumethat Ghas atleastone edge.
4
¯
1 When is A(G) a domain?
In thisparagraph, weanswerthefollowingquestions: When doesA¯(G)contain
– zero-divisors?
– non-zero-divisors?
– nilpotentelements?
The idea, as in Herzog–Hibi–Trung[8], is to use vertex covers as a “bridge” between com-
mutativealgebraand combinatorics:
first we characterize the algebraic properties above with statements in the language of
•
vertexcovers;
then we translate the vertex covers statements into combinatorial properties of the given
•
graphG.
ThiswayweestablishthatA¯(G)isadomainifandonlyifthegraphGhasatleastoneedgeand
satisfiesthefollowing,purely combinatorialcondition.
Definition 1.1. A bipartite graph G satisfies the weak square condition (shortly, WSC) if for
every non-isolated vertex i of G there exists an edge i, j of G such that the following holds
{ }
true:
If i,i isan edgeofG and j, j is an edgeofG,then i, j is alsoan edgeofG.
′ ′ ′ ′
{ } { } { }
It is easy to check that any complete bipartite graph satisfies WSC; furthermore, if two
graphssatisfyWSC, theirdisjointunionsatisfies WSCas well.
Wewillalso provethefollowingprospect:
A¯(G)has anon-zero divisor ifandonlyif G has atleast oneedge;
A¯(G)has azero-divisor ifandonlyif G doesnot satisyWSC;
A¯(G)has anilpotentelement ifandonlyif G has noedges.
Thesecondansweris particularlyinterestingin connectionwiththefollowingresult.
Proposition1.1. SupposethatA¯(G)isadomain. ThenA¯(G)isanormaldomainandaCohen-
Macaulayalgebra,too.
Proof. If A¯(G) is a domain then it is a semigroup ring. More precisely A¯(G) = K[C] where
∼
C Zn+1 is thesemigroup
⊆
C= (a ,...,a ,k): k 1, (a ,...,a ) isabasick-cover .
1 n 1 n
{ ≥ }
By aresultofHochster([12, Theorem1]), itsuffices to showthatC isa normalsemigroup.
Choosean a Z C such thatm a C forsomem N. Write
∈ · · ∈ ∈
p p p
a=((cid:229) m ai, ..., (cid:229) mai, (cid:229) mk )
i 1 i n i i
i=1 i=1 i=1
wherem Z and ai =(ai,...,ai,k) C forevery i=1,...,p. Foreveryedge i, j ofG
i ∈ 1 n i ∈ { }
m ((cid:229) p m (ah+ah)) m ((cid:229) p m k ) 1 = (cid:229) p m (ah+ah) (cid:229) p m k 1,
· h=1 k i j ≥ · h=1 h h ≥ ⇒ h=1 h i j ≥ h=1 h h ≥
andequalityonthelefthandsideimpliesequalityontherightandside. Som a C = a C,
· ∈ ⇒ ∈
thenC isanormalsemigroup.
5
Let usstart withsomeeasy lemmas.
Lemma 1.2. Let G bea bipartitegraph. Then A¯(G)is notadomainifandonlyifthereexistm
basic1-covers thataddup toa non-basicm-cover, forsomem 2.
≥
Proof. Theif-partiseasy.
Conversely, choose two non-zero elements f,g A¯(G) such that f g=0; or, in other words,
∈ ·
choose two elements F,G A(G) mA(G) such that F G= H with H mA(G). Consider a
∈ \ · ∈
term order on S[t]. We can assume that the leading term of F, LT (F), does not belong in
mA(G): oth≺erwise, replace F with F =F a LT F for a suitable a≺ K. Similarly, we can
′
also assume LT (G) / mA(G). On the othe−r ha·nd≺mA(G) is generated∈as a K-vector space by
≺ ∈
monomials,so LT (H) mA(G).
≺ ∈
ThenwemusthaveLT (F) LT (G)=LT (H),so
≺ · ≺ ≺
(xa1 xan tk) (xb1 xbn th)=xc1 xcn tm,
1 ··· n · · 1 ··· n · 1 ··· n ·
wherea isa basick-cover, bis a basich-cover,c is anon basicm-cover, and m=k+h. Find k
basic 1-covers that add up to a, and h basic 1-covers that add up to b. In total we have m basic
1-coversaddingupto somenon-basicm-coverc.
Lemma 1.3. If bisa basick-cover ofa graphG on [n],b k foralli 1,...,n .
i
≤ ∈{ }
Proof. Wheneverb k+1,bcan beloppedat i.
i
≥
Proposition1.4. Let Gbea bipartitegraph. Then, A¯(G) hasnonilpotentelement.
Proof. Ifa isabasick-coverofG andt is anynaturalnumber, thent ais abasic(tk)-coverof
G (by our assumption G has at least one edge). In particular, A¯(G) h·as no nilpotent generator.
Nowarguingas in Lemma1.2wededucethat thereareno nilpotentelementsinA¯(G).
Remark 1.5. Let G be a bipartite graph. Suppose G has isolated vertices. Any basic cover of
G assigns 0 to all the isolated vertices; therefore, A¯(G) is isomorphic as a graded K-algebra to
A¯(G) - where G is the graph obtained deleting from G all the isolated vertices. In particular
′ ′
A¯(G)isadomainifandonly ifA¯(G) isa domain.
′
Theorem 1.6. Let G be a bipartite graph. Then, A¯(G) is a domain if and only if for every
non-isolatedvertexiofGthereexistsanedge i, j ofGsuchthatforeverybasic1-covercone
{ }
hasc +c =1.
i j
Proof. “ ”. By contradiction,ifA¯(G) is not adomain, thereexistm basic1-coversa1, ..., am
⇐
thatadduptoanon-basicm-cover(becauseofLemma1.2). FixavertexiofG;bytheassump-
tion, there is an edge i, j such that ah+ah =1, for all h 1,...,m . But then the m-cover
i j
{ } ∈{ }
a := a1+...+am cannot be lopped at i. Repeat this reasoning for all i to conclude that a is
basic,acontradiction.
“ ”. WecanassumebythepreviousremarkthatGhasnoisolatedpoints. Bycontradiction,
⇒
thereisa vertexisuch thatforeach vertex j adjacent to i,thereis somebasic1-coverb=b(j)
such that b(j) +b(j) =2. (Notice that b(j) +b(j) cannot be 3 or more, because of Lemma
i j i j
1.3.) Nowdefine
(cid:229)
b:= b(j).
j adjacent toi
6
Clearly b is a d-cover, where d >0 is the degree of the vertex i in the graph G. It is also the
sumofd basic1-covers. Itisstraightforwardtocheck that bcan beloppedat i,whichendsthe
proof.
Lemma 1.7. Let i, j bean edgeofa bipartitegraphG. Thefollowingareequivalent:
{ }
(I). a +a =1 foreverybasic1-cover aof G;
i j
(II). if i,i is anedgeand j, j isan edge, then i, j isanedge.
′ ′ ′ ′
{ } { } { }
Proof. (I) (II): By contradiction, suppose i,i , i, j , j, j are edges of G, while i, j
′ ′ ′ ′
⇒ { } { } { } { }
isnotan edge. Wecan define a1-coverbas
0 ifh i, j ,
bh :=(cid:26) 1 othe∈rw{is′e.′}
In general b is not basic; however, we can lop it at vertices not in the set i, j,i, j , until we
′ ′
{ }
obtainabasic1-coverasuch thata +a =2; acontradictionwith(I).
i j
(II) (I): Bytheassumption,thereisanedge i, j suchthatforalli adjacenttoi,andfor
′
⇒ { }
all j adjacent to j, i, j is also an edge of G. By contradiction, suppose there exists a basic
′ ′ ′
{ }
1-coverc of G such that c =c =1. Since c is basic, c cannot be lopped at i, so there must be
i j
somevertex i adjacent to i such that c =0. Similarly, c cannot be lopped at j, so there is a j
′ i ′
′
adjacentto jsuchthatc =0. But i, j isanedgeofG,andc =c =0: acontradiction.
j ′ ′ i j
′ { } ′ ′
Theorem 1.8. Let G bea bipartitegraph. Then thefollowingareequivalent:
1. GsatisfiesWSC;
2. for every non-isolated vertex i of G there exists an edge i, j of G, such that for every
{ }
basic1-cover conehasc +c =1;
i j
3. A¯(G) isa domain.
Proof. By Remark 1.5, it is sufficient to show the equivalence for graphs G without isolated
vertices. The equivalence of 2 and 3 is given by Theorem 1.6; the equivalence of 1 and 2
followsdirectlyby Lemma1.7and thedefinitionofWSC.
Corollary1.9. If GsatisfiesWSC, A¯(G)isa normalCohen-Macaulaydomain.
Examples 1.10. LetGbeasquare. Then,A¯(G)isadomain. IfGisa2k-cycle,andk 3,
• ≥
Gis notadomain.
Let G be (the 1-skeleton of) a cube. Then, A¯(G) is not a domain. However, if G˜ is the
• cubetogetherwithitsthreemaindiagonals,A¯(G˜) isadomain.
IfG isa pathof3 consecutivesegments(i.e. asquareminusan edge), A¯(G)isa domain;
• if G is a path of 4 consecutive segments (i.e. a pentagon minus an edge), A¯(G) is not a
domain.
Recall that G is unmixed if all the basic 1-covers of G have the same number of ones. It
followsfrom the definition that when G is unmixed, m basic 1-covers add up always to a basic
m-cover: so unmixedimpliesdomainby Lemma1.2. Theconversedoes nothold: thealgebras
ofcompletebipartitegraphsare alldomains,whileK is unmixedifandonlyifa=b.
a,b
Proposition1.11. Let T bea treeon nvertices.
T is a domainif andonlyif everyvertexof T iseithera leaf,or adjacenttoa leaf.
•
7
T isunmixedifandonlyifniseven,andT isobtainedfromsometreeU on n verticesby
• 2
attachingoneleafto eachvertex ofU.
The second part of Proposition 1.11 is known: for example, it follows from the result of
Herzog–Hibi[7, Theorem 3.4]and fromthemain theoremby Herzog–Hibi–Zheng[9]. Below,
however,wegiveapurelycombinatorialproof:
Proof. The first item follows applying the WSC property to paths: fix a vertex i and consider
anyedge i, j . Unless i is aleaf (or j is a leaf), wecan find edges i,i and j, j : and since
′ ′
{ } { } { }
atreeisacyclic, i, j is notan edge.
′ ′
{ }
Suppose T is unmixed. Then, the bipartition of the vertices of B gives rise to two basic
1-covers with the same number of ones: so n is even and A = B = n. Moreover, let c be the
| | | | 2
1-cover obtained assigning 0 to all leaves and 1 to all other vertices. A priori c might be not
basic;however,sinceunmixedimpliesdomain,by thepreviousitemeveryvertexofT iseither
a leaf or adjacent to a leaf; this implies that c is basic. Let C be the set of elements that are
adjacent to at least one leaf. Clearly c is identically 1 onC. So the number of elements ofC is
n (by definition of unmixed). Now we notice that every leaf is adjacent to exactly one element
2
inC, while an element v inC is adjacent to d leaves; but since C = n =n # leaves , by
v | | 2 − { }
thepidgeon-holeprincipleweget thatd =1 forallv’s.
v
Conversely, let T be a tree obtained from U as described above, and let b be any basic
1-cover of U. b can be extended in an unique way to a basic 1-cover b of T - and all basic
′
1-covers of T are obtained this way. The extension consists in assigning 0 to a leaf that starts
at a point that has been given 1, and the other way around. This means that the obtained basic
1-coverofT willhaveexactly n ones.
2
Example1.12. LetT bethetreeon sixverticesand thefollowingfiveedges:
1,2 , 1,3 , 1,4 , 4,5 , 4,6 .
{ } { } { } { } { }
Then,T is notunmixed,albeitT isadomain.
Notice that in this case I = (x,x ) = (x x , x x x , x x x , x x x x ). So I is not
i,j i j 1 4 1 5 6 2 3 4 2 3 5 6
generated inonlyonegraden∩ei{the}racompleteintersection,albeititsfiber coneisadomain.
We finally give a criterion to determine whether a basic k-cover is a zero-divisor in the
algebraornot.
Lemma 1.13. Let G be a bipartite graph, and b a basic k-cover. If b +b =k for each edge
i j
i, j ofG, thenthesumof bandother basiccovers givesalways a basiccover.
{ }
Proof. Let a=b+c be the sum of the basick-cover b aboveand an arbitrary basic k -coverc.
′
aisa(k+k )-cover;weclaimthata cannotbeloppedatanyvertex. By contradiction,suppose
′
a can be lopped at i: then, the inequality a +a k+k was strict for all j’s adjacent to i.
j i ′
≥
Besides,b +b =k forall j’sadjacent toi. So,c +c k mustbestrictforall j’sadjacent to
i j j i ′
≥
i-which meansthat ccan beloppedat i,acontradiction.
Proposition1.14. Let G bea bipartitegraphwith atleast oneedge. Then depth A¯(G) 2.
≥
(cid:0) (cid:1)
8
Proof. By Remark 1.5 we can assume that G has no isolated vertices. Under this assumption,
thevectorsa and bdefined by
1 ifh A,
ah :=(cid:26) 0 ifh∈B
∈
0 ifh A,
bh :=(cid:26) 1 ifh∈B
∈
arebothbasick-covers,where[n]=A Bis thebipartitionofthevertexset ofG.
∪
We claim that xat =xa1 xant and xbt =xb1 xbnt form a regular sequence in the algebra
A¯(G). 1 ··· n 1 ··· n
In fact, by Lemma 1.13, the sum of a and other basic covers gives always a basic cover,
so xat is a non-zero divisor arguing like in the proof of Lemma 1.2. To prove that xbt is not
a zero-divisor of A¯(G)/xat A¯(G) we assume xbt f (x ,...,x ,xat)A¯(G) and we show f
1 n
(x ,...,x ,xat)A¯(G) (here f· A(G) and the opera·tion∈s are in S). Again, arguing as in Lemm∈a
1 n
1.2 we can assume that f =∈xc1 xcntk where c = (c ,...,c ) is a basic k-cover. Moreover,
1 ··· n 1 n
since xbt is not a zero-divisor(again by Lemma 1.13) we have that there exists a basic k-cover
d such that a+d=b+c. Now it is easy to see that, by the choice of a and b, this implies that
c=a+c fora suitablebasic(k 1)-coverc, and thisallowsus toconclude.
′ ′
−
Corollary 1.15. Let G be a bipartite graph. Then the projective scheme Proj(A¯(G)) is con-
nected.
Proof. From a result of Hartshorne (see [6, Proposition 2.1]) one has that for any projective
schemeX =Proj(R)thefollowingfact holds: depth(R) 2 = X connected.
≥ ⇒
Theorem 1.16 (Characterization of non-zero-divisors.). Let G be a bipartite graph, and b a
basick-cover. Thefollowingareequivalent:
1. thesum ofband otherbasiccovers gives alwaysa basiccover;
2. for each edge i, j of G such that b +b > k, there exist an i adjacent to i and a j
i j ′ ′
{ }
adjacentto j suchthatforeach basic1-cover a, onehas a +a =1=a +a ;
i i j j
′ ′
3. for each edge i, j of G such that b +b > k, there exist an i adjacent to i and a j
i j ′ ′
{ }
adjacentto j suchthat:
– if v,i is anedgeand v,i isan edge, then v,v is alsoanedge;
′ ′ ′
{ } { } { }
– if w, j isan edgeand w, j is anedge, then w,w isalsoan edge.
′ ′ ′
{ } { } { }
Proof. Theequivalenceof2 and 3is easilyshownviaLemma1.7.
1 2: By contradiction, let i, j be an edge with b +b >k, and suppose that for all i
i j ′
⇒ { }
adjacent to i and j adjacent to j, onecan find eithera basic1-coverc=c(i) (dependingon i)
′ ′ ′
suchthat c +c =2,orabasic1-coverd=d(j ) (dependingon j )such thatd +d =2.
i i ′ ′ j j
′ ′
Consider
(cid:229) (cid:229)
a:= c(i ) + d(j ).
′ ′
i adjacent toi j adjacent to j
′ ′
Thisa isa h-cover,where h is thesumof thedegrees ofi and j inthegraph G. It iseasy to
seethata can beloppedeitherat iorat j.
2 1: Bycontradiction,letcbeabasick -coversuchthata=b+cisnotbasic. Reasoning
′
⇒
as in Lemma 1.13, it is easy to see that c could be lopped only at those vertices i such that
b +b > k for some j adjacent to i. So, fix such an edge i, j with b +b > k, so that c
i j i j
{ }
9
can be lopped at i. Clearly c is a sum of k indecomposable1-covers; by the assumption, then,
′
c +c =k =c +c ,forsomei adjacenttoi,andsome j adjacentto j. Butifccanbelopped
i i ′ j j ′ ′
′ ′
ati, theinequalityc +c k shouldbestrict: acontradiction.
i i ′
′ ≥
¯
2 The dimension of A(G)
In thisparagraph westudythedimensionofA¯(G),whichis theanalyticspread oftheideal
I = (x,x ) S=K[x ,...,x ].
i j 1 n
⊆
i,j ed\geofG
{ }
As wehavealready said theHilbertfunctionofA¯(G)
HFA¯(G)(k)=dimKA¯(G)k
counts the number of basic k-covers of G. For k large, the Hilbert function grows as C kd,
where d =dimA¯(G) 1 andC d!=e(A¯(G)), the multiplicity of A¯(G). Hence the main·tool
− ·
for computing the dimension will be approximating the number of basic k-cover of the graph
fork large.
We introduceherein the“graphical dimension”,an integerthat depends only on thecombi-
natoricsofG. Thisinvariantcaptured ourinterestforthefollowingreasons:
⊲ gdim(G) dimA¯(G)(Theorem 2.7);
≤
⊲ gdimcan becomputedorestimatedfrombelow;2
⊲ gdim(C )=dimA¯(C )=aforeach (2a)-gon(see Proposition2.12);
2a 2a
⊲ gdim(G) = dimA¯(G) = n +1 if G is a path on n vertices, or if G has a Cohen-
⌊2⌋
Macaulay edgeideal(see Corollaries2.8 and 2.9);
⊲ gdim(T)=dimA¯(T)foreach treeT (Theorem 2.22).
Notice that by Remark 1.5, we may assume that G has no isolated points. We can even reduce
tothecasewhereG is connectedby meansofthefollowingLemma:
Lemma 2.1. If G isthedisjointunionof two graphsG andG ,
1 2
dim(A¯(G))=dim(A¯(G ))+dim(A¯(G )) 1.
1 2
−
Proof. Every basic k-cover of G, restricted to G , gives a basic k-cover of G ; and all basic
i i
k-coversofGarisethisway. Thismeansthat
HF (k) = HF (k) HF (k),
A¯(G) A¯(G1) · A¯(G2)
whenceforlargek weobtainthat
dim(A¯(G)) 1 = dim(A¯(G )) 1 + dim(A¯(G )) 1.
1 2
− − −
2Computinggdimtakes“toolong”ifGhasmanyedges. However,anydrawingthatrepresentsGasbipartite
graphbasicallygivesalowerboundforgdim(G),andthusfordimA¯(G): SeeDefinition2.1.
10