Table Of ContentA class of transversal polymatroids with Gorenstein base ring
Alin S¸tefan
”Petroleum and Gas” University of Ploie¸sti, Romania
Abstract
8
0 Inthispaper,theprincipaltooltodescribetransversalpolymatroidswithGorensteinbase
0 ringispolyhedralgeometry,especiallytheDanilov−Stanley theoremforthecharacterization
2 of canonical module. Also, we compute the a−invariant and theHilbert series of base ring
n associated to this class of transversal polymatroids.
a
J
1 Introduction
5
1
In this paper we determine the facets of the polyhedral cone generated by the exponent set of
]
C the monomials defining the base ring associated to a transversal polymatroid. The importance
of knowing those facets comes from the fact that the canonical module of the base ring can
A
be expressed in terms of the relative interior of the cone. This would allow to compute the
.
h a−invariant of those base rings. The results presented were discovered by extensive computer
t algebra experiments performed with Normaliz [4].
a
m TheauthorwouldliketothankProfessorDorinPopescuforvaluablesuggestionsandcomments
during the preparation of this paper.
[
1
v 2 Preliminaries
9
2
3
2 Let n ∈ N, n ≥ 3, σ ∈ Sn, σ = (1,2,...,n) the cycle of length n, [n] := {1,2,...,n} and
1. {ei}1≤i≤n be the canonicalbase of Rn. For a vector x∈Rn, x=(x1,...,xn) we will denote by
0 | x |, | x |:=x +...+x . If xa is a monomial in K[x ,...,x ] we set log(xa)=a. Given a set
1 n 1 n
8 A of monomials, the log set of A, denoted log(A), consists of all log(xa) with xa ∈A.
0
We consider the following set of integer vectors of Nn:
:
v
Xi ↓ithcolumn
r ν := −(n−i−1), −(n−i−1), ... ,−(n−i−1), (i+1), ... ,(i+1) ,
a σ0[i]
(cid:0) (cid:1)
↓(2)thcolumn ↓(i+1)thcolumn
ν := (i+1), −(n−i−1), ... ,−(n−i−1), (i+1), ... ,(i+1) ,
σ1[i]
(cid:0) (cid:1)
↓(3)thcolumn ↓(i+2)thcolumn
ν := (i+1), (i+1) ,−(n−i−1), ... ,−(n−i−1), (i+1), ... ,(i+1) ,
σ2[i]
(cid:0) (cid:1)
..........................................................................................
↓(i−2)thcolumn ↓(n−2)thcolumn
νσn−2[i] := −(n−i−1),...,−(n−i−1),(i+1), ... ,(i+1),−(n−i−1),−(n−i−1) ,
(cid:0) (cid:1)
1
↓(i−1)thcolumn ↓(n−1)thcolumn
νσn−1[i] := −(n−i−1), ... ,−(n−i−1),(i+1), ... ,(i+1),−(n−i−1) ,
(cid:0) (cid:1)
where σk[i]:={σk(1),...,σk(i)} for all 1≤i≤n−1 and 0≤k ≤n−1.
Remark: ν =n e for all 0≤k ≤n.
σk[n−1] [n]\σk[n−1]
For example, if n=4, σ =(1,2,3,4)∈S then we have the following set of integer vectors:
4
ν =ν =(−2,2,2,2), ν =ν =(−1,−1,3,3), ν =ν =(0,0,0,4),
σ0[1] {1} σ0[2] {1,2} σ0[3] {1,2,3}
ν =ν =(2,−2,2,2), ν =ν =(3,−1,−1,3), ν =ν =(4,0,0,0),
σ1[1] {2} σ1[2] {2,3} σ1[3] {2,3,4}
ν =ν =(2,2,−2,2), ν =ν =(3,3,−1,−1), ν =ν =(0,4,0,0),
σ2[1] {3} σ2[2] {3,4} σ2[3] {1,3,4}
ν =ν =(2,2,2,−2), ν =ν =(−1,3,3,−1), ν =ν =(0,0,0,4).
σ3[1] {4} σ3[2] {1,4} σ3[3] {1,2,4}
If06=a∈Rn,thenHa willdenotethehyperplaneofRn throughtheoriginwithnormalvector
a, that is,
Ha ={x∈Rn | <x,a>=0},
where <,> is the usual inner product in Rn. The two closed halfspaces bounded by Ha are:
H+ ={x∈Rn | <x,a>≥0} and H− ={x∈Rn | <x,a>≤0}.
a a
We will denote by H the hyperplane of Rn through the origin with normal vector ν ,
σk[i] σk[i]
that is,
Hνσk[i] ={x∈Rn | <x,νσk[i] >=0},
for all 1≤i≤n−1 and 0≤k ≤n−1.
Recallthatapolyhedral coneQ⊂Rn istheintersectionofafinitenumberofclosedsubspaces
of the form Ha+. If A = {γ1,..., γr} is a finite set of points in Rn the cone generated by A,
denoted by R A, is defined as
+
r
R+A={ aiγi | ai ∈R+, with 1≤i≤n}.
Xi=1
An important fact is that Q is a polyhedral cone in Rn if and only if there exists a finite set
A⊂Rn such that Q=R A, see ([3] [10],Theorem 4.1.1.).
+
Next we give some important definitions and results.(see [1], [2], [3], [8], [9].)
Definition 2.1. A proper face of a polyhedral cone is a subset F ⊂ Q such that there is a
supporting hyperplane H satisfying:
a
1) F =Q∩H 6=∅.
a
2) Q*Ha and Q⊂Ha+.
Definition 2.2. AconeC is apointedif0isa faceofC. Equivalentlywecanrequirethatx∈C
and −x∈C ⇒ x=0.
Definition 2.3. The 1-dimensional faces of a pointed cone are called extremal rays.
Definition 2.4. If a polyhedral cone Q is written as
Q=H+ ∩...∩H+
a1 ar
such that no one of the H+ canbe omitted, then we saythat this is anirreducible representation
ai
of Q.
2
Definition 2.5. AproperfaceF ofapolyhedralconeQ⊂ Rn iscalledafacetofQifdim(F)=
dim(Q)−1.
Definition 2.6. Let Q be a polyhedralcone in Rn withdim Q=n andsuchthat Q6=Rn. Let
Q=H+ ∩...∩H+
a1 ar
be the irreducible representation of Q. If a =(a ,...,a ), then we call
i i1 in
H (x):=a x +...+a x =0,
ai i1 1 in n
i∈[r], the equations of the cone Q.
The following result gives us the description of the relative interior of a polyhedral cone when
we know the irreducible representation of it.
Theorem 2.7. Let Q⊂Rn, Q6=Rn be a polyhedral cone with dim(Q)=n and let
(∗) Q=H+ ∩...∩H+
a1 an
be a irreducible representation of Q with Ha+1,...,Ha+n distinct, where ai ∈ Rn\{0} for all i.
Set F =Q∩H for i∈[r]. Then :
i ai
a) ri(Q)={x∈Rn | <x,a1 > >0,...,<x,ar > >0}, where ri(Q) is the relative interior of
Q, which in this case is just the interior.
b) Each facet F of Q is of the form F =F for some i.
i
c) Each F is a facet of Q if and only if (∗) is irreducible.
i
Proof. See [1] Theorem 8.2.15, Theorem 3.2.1.
Theorem 2.8. (Danilov, Stanley) Let R = K[x ,...,x ] be a polynomial ring over a field K
1 n
andF afinitesetof monomials inR.IfK[F]is normal, thenthecanonicalmoduleω ofK[F],
K[F]
with respect to standard grading, can be expressed as an ideal of K[F] generated by monomials
ω =({xa| a∈NA∩ri(R A)}),
K[F] +
where A=log(F) and ri(R A) denotes the relative interior of R A.
+ +
The formula above represents the canonical module of K[F] as an ideal of K[F] generated
by monomials. For a comprehensive treatment of the Danilov−Stanley formula see [2], [8] [9].
3 Polymatroids
Let K be a infinite field, n and m be positive integers, [n] = {1,2,...,n}. A nonempty
finite set B of Nn is the base set of a discrete polymatroid P if for every u = (u ,u ,...,u ),
1 2 n
v = (v ,v ,...,v ) ∈ B one has u +u +...+u = v +v +...+v and for all i such that
1 2 n 1 2 n 1 2 n
u >v thereexistsj suchthatu <v andu+e −e ∈B,wheree denotesthe kth vectorofthe
i i j j j i k
standardbasisofNn. Thenotionofdiscretepolymatroidisageneralizationoftheclassicalnotion
of matroid, see [5] [6] [7] [11]. Associated with the base B of a discret polymatroid P one has a
K−algebra K[B], called the base ring of P, defined to be the K−subalgebra of the polynomial
ring in n indeterminates K[x ,x ,...,x ] generated by the monomials xu with u ∈ B. From [7]
1 2 n
the algebra K[B] is known to be normal and hence Cohen-Macaulay.
If A are some non-empty subsets of [n] for 1 ≤ i ≤ m, A = {A ,...,A }, then the set of
i 1 m
m
the vectors e with i ∈ A , is the base of a polymatroid, called transversal polymatroid
k=1 ik k k
presented byPA. The base ring of a transversal polymatroid presented by A denoted by K[A] is
the ring :
K[A]:=K[x ···x :i ∈A ,1≤j ≤m].
i1 im j j
3
4 Cones of dimension n with n + 1 facets.
Lemma 4.1. Let 1 ≤ i ≤ n−2, A := {log(xj1 ···xjn) | jk ∈ Ak, for all 1 ≤ k ≤ n} ⊂ Nn the
exponent set of generators of K−algebra K[A], where A = {A = [n],...,A = [n],A = [n]\
1 i i+1
[i],...,A =[n]\[i],A =[n]}. Then theconegeneratedbyAhastheirreduciblerepresentation:
n−1 n
R A= H+,
+ a
a\∈N
where N ={ν ,ν | 0≤k ≤n−1}.
σ0[i] σk[n−1]
(i+1) e +(n−i−1) e , if 1≤k ≤i
Proof. WedenotebyJ = k i+1 andbyJ =ne .
k (cid:26) (i+1) e1+(n−i−1) ek , if i+2≤k ≤n n
Since A =[n] for any t∈{1,...,i}∪{n} and A =[n]\[i] for any r ∈{i+1,...,n−1} it is
t r
easy to see that for any k ∈{1,...,i} and r ∈{i+2,...,n} the set of monomials xi+1 xn−i−1,
k i+1
xi+1 xn−i−1, xn are a subset of the generators of K−algebra K[A]. Thus the set
1 r n
{J ,...,J ,J ,...,J ,J}⊂A.
1 i i+2 n
If we denote by C the matrix with the rows the coordinates of the {J ,...,J ,J ,...,J ,J},
1 i i+2 n
then by a simple computationwe get|det (C)|=n (i+1)i (n−i−1)n−i−1 for any1≤i≤n−2
and 1≤j ≤n−1. Thus, we get that the set
{J ,...,J ,J ,...,J ,J}
1 i i+2 n
islinearyindependentanditfollowsthatdimR+A=n. Since{J1,...,Ji,Ji+2,...,Jn}islinearly
independent and lie on the hyperplane H we have that dim(H ∩R A)=n−1.
σ0[i] σ0[i] +
Now we will provethat R A⊂H+ for all a∈N. It is enoughto show that for all vectorsP ∈A,
+ a
<P,a >≥0 for all a∈N. Since ν =n e , where {e } is the canonical base
σk[n−1] [n]\σk[n−1] i 1≤i≤n
ofRn,we getthat <P,νσk[n−1] >≥0. LetP ∈A, P =log(xj1···xjixji+1···xjn−1xjn) andlet t
to be the number of j , such that 1≤k ≤i and j ∈[i]. Thus 1≤t≤i. Now we have only
ks s ks
two cases to consider:
1) If j ∈[i], then <P,ν >=−t(n−i−1)+(i−t)(i+1)+(n−i−1)(i+1)−(n−i−1)=
n σ0[i]
n(i−t)≥0.
2) If j ∈[n]\[i], then <P,ν >=−t(n−i−1)+(i−t)(i+1)+(n−i−1)(i+1)+(i+1)=
n σ0[i]
n(i−t+1)>0.
Thus
R A⊆ H+.
+ a
a\∈N
Now we will prove the converse inclusion: R A⊇ H+.
+ a∈N a
It is clearly enough to prove that the extremal raTys of the cone a∈NHa+ are in R+A. Any
extremal ray of the cone a∈NHa+ can be written as the intersectTion of n−1 hyperplanes Ha ,
witha∈N. TherearetwoTpossibilitiestoobtainextremalraysbyintersectionofn−1hyperplanes.
First case.
Let 1≤i <...<i ≤n beasequenceand{t}=[n]\{i ,...,i }. Thesystemofequations:
1 n−1 1 n−1
zi1 =0, x1
(∗) ... admits the solution x∈Zn+, x= ... with | x |=n, xk =n·δkt for all
zin−1 =0 xn
1≤k ≤n, where δkt is Kronecker symbol.
There are two possibilities:
1) If 1≤t≤i, then H (x) < 0 and thus x ∈/ H+.
σ0[i] a∈N a
2) If i+1≤t≤n, then Hσ0[i](x) > 0 and thus xT∈ a∈NHa+ and is an extremal ray.
T
4
Thus, there exist n−i sequences 1≤i <...<i ≤n such that the system of equations (∗)
1 n−1
has a solution x∈Zn+ with | x |=n and Hσ0[i](x) > 0.
The extremal rays are: {ne | i+1≤k ≤n}.
k
Second case.
Let 1 ≤ i < ... < i ≤ n be a sequence and {j, k} = [n]\{i ,...,i }, with j < k and
1 n−2 1 n−2
z =0,
i1
.
.
(∗∗) .z =0,
in−2
−(n−i−1)z −...−(n−i−1)z +(i+1)z +...+(i+1)z =0
be the system of line1ar equations associateid to this sei+q1uence. n
There are two possibilities:
1) If 1 ≤ j ≤ i and i + 1 ≤ k ≤ n, then the system of equations (∗∗) admits the solution
x
1
x= ... ∈Zn+, with | x |=n, with xt =(i+1)δjt+(n−i−1)δkt for all 1≤t≤n.
xn
2) If 1≤j,k ≤i or i+1≤j,k ≤n, then there exist no solution x∈Zn with | x | = n for the
+
system of equations (∗∗) because otherwise H (x) > 0 or H (x)<0.
σ0[i] σ0[i]
Thus, there exist i(n−i) sequences 1 ≤ i < ... < i ≤ n such that the system of equations
1 n−2
(∗∗) hasasolutionx∈Zn+ with|x|=n andtheextremalraysare: {(i+1)ej+(n−i−1)ek|1≤
j ≤i and i+1≤k ≤n}.
In conclusion, there exist (i+1)(n−i) extremal rays of the cone H+:
a∈N a
T
R:={ne | i+1≤k ≤n}∪{(i+1)e +(n−i−1)e | 1≤j ≤i and i+1≤k ≤n}.
k j k
Since R⊂A we have R A= H+.
+ a∈N a
It is easy to see that the repTresentation is irreducible because if we delete, for some k, the
hyperplanewith the normalν ,then a coordinateofa log(x ···x x ···x x )could
σk[n−1] j1 ji ji+1 jn−1 jn
be negative, which is impossible; and if we delete the hyperplane with the normal ν , then the
σ0[i]
cone R+A would be generated by A = {log(xj1 ···xjn) | jk ∈ [n], for all 1 ≤ k ≤ n} which is
impossible. Thus the representation R A= H+ is irreducible.
+ a∈N a
T
Lemma 4.2. Let 1 ≤ i ≤ n−2, 1 ≤ t ≤ n−1, A := {log(x ···x ) | j ∈ A ,1 ≤
j1 jn σt(k) σt(k)
k ≤ n} ⊂ Nn the exponent set of generators of K−algebra K[A], where A = {Aσt(k) | Aσt(k) =
[n], for 1≤k ≤i and A =[n]\σt[i], for i+1≤k ≤n−1, A =[n]}. Then the cone
σt(k) σt(n)
generated by A has the irreducible representation:
R A= H+,
+ a
a\∈N
where N ={ν , ν | 0≤k ≤n−1}.
σt[i] σk[n−1]
Proof. The proof goes as in Lemma 4.1. since the algebras from Lemmas 4.1. and 4.2. are
isomorphic.
5 The a-invariant and the canonical module
Lemma5.1. TheK−algebraK[A],whereA={A |A =[n], for1≤k ≤i, andA =
σt(k) σt(k) σt(k)
[n]\σt[i], for i+1 ≤ k ≤ n−1, A = [n]}, is Gorenstein ring for all 0 ≤ t ≤ n−1 and
σt(n)
1≤i≤n−2.
Proof. Since the algebras from Lemmas 4.1 and 4.2 are isomorphic it is enough to prove the case
t=0.
5
We will show that the canonical module ω is generated by (x ···x )K[A]. Since K−
K[A] 1 n
algebra K[A] is normal, using the Danilov−Stanley theorem we get that the canonical module
ω is
K[A]
ω ={xα | α∈NA∩ri(R A)}.
K[A] +
Let d be the greatest common divizor of n and i+1, d=gcd(n, i+1), then the equation of the
facet H is
νσ0[i]
i n
(n−i−1) (i+1)
H : − x + x =0.
νσ0[i] d k d k
kX=1 k=Xi+1
The relative interior of the cone R A is:
+
i n
(n−i−1) (i+1)
ri(R+A)={x∈Rn | xk > 0, for all 1≤k ≤n, − xk+ xk > 0}.
d d
Xk=1 k=Xi+1
We will show that NA∩ri(R A)= (1,...,1) + (NA∩R A).
+ +
It is clear that ri(R A) ⊃ (1,...,1) +R A.
+ +
If (α1,α2,...,αn)∈NA∩ri(R+A), then αk ≥1,for all 1≤k ≤n and
i n n
(n−i−1) (i+1)
− α + α ≥1 and α =t n for some t≥1.
k k k
d d
kX=1 k=Xi+1 kX=1
We claim that there exist (β1,β2,...,βn) ∈ NA∩R+A such that (α1,α2,...,αn) = (β1 +
1,β +1,...,β +1). Let β =α −1 for all 1≤k ≤n. It is clear that β ≥0 and
2 n k k k
i n i n
(n−i−1) (i+1) (n−i−1) (i+1) n
− β + β =− α + α − .
k k k k
d d d d d
kX=1 k=Xi+1 kX=1 k=Xi+1
i n
(n−i−1) (i+1) n
If− α + α =j with1≤j ≤ −1, thenwewillgetacontadiction.
k k
d d d
Xk=1 k=Xi+1
Indeed, since n divides n α , it follows n divides j which is false.
k=1 k d
So we have P
i n i n
(n−i−1) (i+1) (n−i−1) (i+1) n
− β + β =− α + α − ≥0.
k k k k
d d d d d
kX=1 k=Xi+1 kX=1 k=Xi+1
Thus (β1,β2,...,βn)∈NA∩R+A and (α1,α2,...,αn)∈NA∩ri(R+A).
Since NA∩ri(R+A)= (1,...,1) + (NA∩R+A), we get that ωK[A] = (x1···xn)K[A].
Let S be a standard graded K −algebra over a field K. Recall that the a−invariant of
S, denoted a(S), is the degree as a rational function of the Hilbert series of S, see for instance
([9], p. 99). If S is Cohen−Macaulay and ω is the canonical module of S, then
S
a(S)= − min {i | (ω ) 6=0},
S i
see ([2], p. 141) and ([9], Proposition 4.2.3). In our situation S = K[A] is normal [7] and
consequently Cohen−Macaulay, thus this formula applies. As consequence of Lemma 5.1. we
have the following:
Corollary 5.2. The a−invariant of K[A] is a(K[A])= −1.
Proof. Let {xα1,...,xαq} the generators of K−algebra K[A]. K[A] is standard graded algebra
with the grading
K[A]i = K(xα1)c1 ···(xαq)cq, where |c|=c1+...+cq.
|Xc|=i
Sinceω = (x ···x )K[A]itfollowsthatmin{i|(ω ) 6=0}=1,thusa(K[A])= −1.
K[A] 1 n K[A] i
6
6 Ehrhart function
We consider a fixed set of distinct monomials F = {xα1,...,xαr} in a polynomial ring R =
K[x ,...,x ] over a field K.
1 n
Let
P =conv(log(F))
be the convex hull of the set log(F)={α ,...,α }.
1 r
The normalized Ehrhart ring of P is the graded algebra
∞
A = (A ) ⊂R[T]
P P j
Mj=0
where the j−th component is given by
(A ) = K xα Tj.
P j
α∈Z loXg(F)∩ jP
The normalized Ehrhart function of P is defined as
EP(j)=dimK(AP)j =| Z log(F)∩jP |.
From [9], Proposition7.2.39 and Corollary 7.2.45 we have the following important result:
Theorem 6.1. If K[F] is a standard graded subalgebra of R and h is the Hilbert function of
K[F], then:
a) h(j)≤E (j) for all j ≥0, and
P
b) h(j)=E (j) for all j ≥0 if and only if K[F] is normal.
P
In this section we will compute the Hilbert function and the Hilbert series for K− algebra
K[A], where A satisfied the hypothesis of Lemma 4.1.
Proposition 6.2. In the hypothesis of Lemma 4.1., the Hilbert function of K− algebra K[A] is :
(i+1)t
k+i−1 nt−k+n−i−1
h(t)= .
(cid:18) k (cid:19)(cid:18) nt−k (cid:19)
Xk=0
Proof. From [7] we know that the K− algebra K[A] is normal. Thus, to compute the Hilbert
function of K[A] it is equivalent to compute the Ehrhart function of P, where P = conv(A).
ItisclearenoughthatP istheintersectionoftheconeR+Awiththehyperplanex1+...+xn =n,
thus
P ={α∈Rn | αk ≥0 for any k ∈[n], 0≤α1+...+αi ≤i+1 and α1+...+αn =n}
and
t P ={α∈Rn | αk ≥0 for any k∈[n], 0≤α1+...+αi ≤(i+1) t and α1+...+αn =n t}.
Since for any 0 ≤ k ≤ (i + 1) t the equation α + ...+ α = k has k+i−1 nonnegative
1 i k
integersolutionsandthe equationα +...+α =n t−k has nt−k+n−i−1(cid:0)nonne(cid:1)gativeinteger
i+1 n nt−k
solutions, we get that (cid:0) (cid:1)
(i+1)t
k+i−1 nt−k+n−i−1
E (t)=| Z A ∩ t P |= .
P
(cid:18) k (cid:19)(cid:18) nt−k (cid:19)
kX=0
7
Corollary 6.3. The Hilbert series of K− algebra K[A], where A satisfied the hypothesis of
Lemma 4.1. is :
1+h t+...+h tn−1
1 n−1
H (t) = ,
K[A] (1−t)n
where
j
n
h = (−1)s h(j−s) ,
j
(cid:18)s(cid:19)
Xs=0
h(s) is the Hilbert function of K[A].
Proof. Since the a−invariant of K[A] is a(K[A])= −1, it follows that to compute the Hilbert
series of K[A] is necessary to know the first n values of the Hilbert function of K[A], h(i) for
0 ≤ i ≤ n−1. Since dim(K[A]) = n, apllying n times the difference operator ∆ (see [2]) on
the Hilbert function of K[A] we get the conclusion.
Let ∆0(h) :=h(j) for any 0≤j ≤n−1.
j
For k ≥1 let ∆k(h) :=1 and ∆k(h) :=∆k−1(h) −∆k−1(h) for any 1≤j ≤n−1.
0 j j j−1
We claim that:
k
k
∆k(h) = (−1)sh(j−s)
j
(cid:18)s(cid:19)
Xs=0
for any k ≥1 and 0≤j ≤n−1.
We proceed by induction on k.
If k =1, then
1
1
∆1(h) =∆0(h) −∆0(h) =h(j)−h(j−1)= (−1)sh(j−s)
j j j−1
(cid:18)s(cid:19)
Xs=0
for any 1≤j ≤n−1.
If k >1, then
k−1 k−1
k−1 k−1
∆k(h) =∆k−1(h) −∆k−1(h) = (−1)sh(j−s) − (−1)sh(j−1−s) =
j j j−1
(cid:18) s (cid:19) (cid:18) s (cid:19)
Xs=0 Xs=0
k−1 k−2
k−1 k−1 k−1 k−1
=h(j) + (−1)sh(j−s) − (−1)sh(j−1−s) +(−1)kh(j−k) =
(cid:18) 0 (cid:19) (cid:18) s (cid:19) (cid:18) s (cid:19) (cid:18)k−1(cid:19)
Xs=1 Xs=0
k−1
k−1 k−1 k−1
=h(j)+ (−1)sh(j−s) + +(−1)kh(j−k) =
(cid:20)(cid:18) s (cid:19) (cid:18)s−1(cid:19)(cid:21) (cid:18)k−1(cid:19)
Xs=1
k−1 k
k k−1 k
=h(j)+ (−1)sh(j−s) +(−1)kh(j−k) = (−1)sh(j−s) .
(cid:18)s(cid:19) (cid:18)k−1(cid:19) (cid:18)s(cid:19)
Xs=1 Xs=0
Thus, if k =n it follows that:
n j
n n
h =∆n(h) = (−1)sh(j−s) = (−1)sh(j−s)
j j
(cid:18)s(cid:19) (cid:18)s(cid:19)
Xs=0 Xs=0
for any 1≤j ≤n−1.
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References
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affine semigroups, 1998. Available via anonymous ftp from: ftp.mathematik.Uni-
Osnabrueck.DE/pub/osm/kommalg/software.
[5] J Edmonds, Submodular functions, matroids, and certain polyedra, in Combinatorial Struc-
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[6] J. Oxley, Matroid Theory, Oxford University Press, Oxford, 1992.
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[8] E.MillerandBSturmfels,Combinatorial commutative algebra, GraduateTextsinMathemat-
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[10] R. Webster, Convexity, Oxford University Press, Oxford, 1994.
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AlinS¸tefan,AssistantProfessor
”PetroleumandGas”UniversityofPloie¸sti
Ploie¸sti,Romania
E-mail:[email protected]
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