Table Of ContentON THE GENERALIZED SCARF COMPLEX
OF LATTICE IDEALS
9
HARACHARALAMBOUSANDAPOSTOLOSTHOMA
0
0
2 Abstract. Letkbeafield,L⊂Zn bealatticesuchthatL∩Nn={0},and
n IL⊂k[x1,...,xn]thecorrespondinglatticeideal. Wepresentthegeneralized
a Scarf complex of IL and show that it is indispensable in the sense that it is
J contained ineveryminimalfreeresolutionofR/IL.
9
]
C 1. Introduction
A
Let k be a field, let L ⊂ Zn be a lattice and I the corresponding lattice ideal
. L
h ink[x ,...,x ]. In a seriesof recentpapers,see [1], [5], [9], [10], [13], [16], [17] and
1 n
t [25], the problem of getting different minimal generating systems for lattice ideals
a
m I was investigated. One of the motivations was a related question from Algebraic
L
Statistics that asked when a prime lattice ideal possesses a unique minimal system
[
of binomial generators. As a result the notion of indispensable binomials was in-
1
troduced by Ohsugi and Hibi [16] to describe the binomials that up to constant
v
multiples are part of all minimal systems of generators of I . The corresponding
8 L
0 notion for higher syzygies is the motivating question for the present article. Three
2 landmarkpapers dealing with free resolutions oflattice ideals by Bayer,Peeva and
1 Sturmfels providecrucialleads: see [4], [20], and[21]. The algebraicScarfcomplex
.
1 isdefinedin[20]anditisshowntobe aminimalfreeresolutionwhenIL isgeneric.
0 TheTaylorcomplexdefinedin[4]isalwaysaresolution,alashardlyeveraminimal
9 one. The complexity of computing minimal resolutions becomes even more appar-
0
ent in [21], and the importance ofthe topologicalstructureof the fibers is strongly
:
v hinted. Apart from the above, the problem of computing syzygies of lattice ideals
i has also been addressed by many recent articles, see for example [6], [8], [18], [19].
X
The structure of this paper is as follows: in section 2 we introduce the basic
r
a notions and terminology. In section 3 we introduce simple minimal resolutions of
I : these are complexes such that the syzygies determined by Zn/L-homogeneous
L
baseshaveminimalsupport. Insection4wegeneralizetheScarfsimplicialcomplex
and basic fibers of [20]. In Section 5 we introduce the generalized algebraic Scarf
complex and show that it is an indispensable complex. The last section contains
examples and open problems.
2. Notation
Let L ⊂ Zn be a lattice such that L∩Nn = {0}. The polynomial ring R =
k[x ,...,x ] is positively multigraded by the group Zn/L, see [15]. Let a = e +
1 n i i
L where {e : 1 ≤ i ≤ n} is the canonical basis of Zn. By A we denote the
i
1991 Mathematics Subject Classification. 13D02,13D25.
Key words and phrases. Resolutions, lattice ideal, syzygies, indispensable syzygies, Scarf
complex.
1
2 H.CHARALAMBOUSANDA.THOMA
subsemigroupofthegroupZn/Lgeneratedby{a :1≤i≤n}. SinceL∩Nn ={0},
i
the semigroup A is pointed meaning that {x : x ∈ A and − x ∈ A} = {0}.
Equivalently if A is a pointed semigroup generated by a ,...,a then L ⊂ Zn is
1 n
the lattice of relations of a ,...,a and L∩Nn ={0}. We set deg (x )=a . The
1 n A i i
A-degree of the monomial xv =xv1···xvn is
1 n
v
deg (x ):=v a +···+v a ∈A.
A 1 1 n n
When we want to put the emphasis on L we occasionally write deg (xv) :=
L
deg (xv). The lattice ideal I (or I ), associated to L is the ideal generated
A L A
by all the binomials xu+ −xu− where u+,u− ∈Nn and u=u+−u− ∈L. Prime
latticeidealsarecalledtoricideals[24]andarethe definingidealsoftoricvarieties.
For binomials in IL, we define degA(xu+ −xu−) := degAxu+. Lattice ideals are
A-homogeneous. For b ∈ A we let R[−b] be the A-graded free R-module of rank
1 whose generator has A-degree b. Let
(F ,φ): 0−→F −φ→p ······−→F −φ→1 F −→R/I −→0,
L p 1 0 L
be a minimal A-graded free resolution of R/I . The i-Betti number of R/I of
L L
A-degree b is equal to the rank of the R-summand of F of A-degree b:
i
βi,b(R/IL)=dimkTori(R/IL,k)b
and is denoted by βi,b(R/IL). This is an invariant of IL, see [15]. For b ∈ A, we
let Cb equal the fiber
Cb :=deg−A1(b)=deg−L1(b):={xu : degA(xu)=b}
and
a a
∆b :={F ⊂{1,...,n}: ∃ x ∈Cb, F ⊂suppx }.
It is well known that
βi,b(R/IL)=dimkH˜i(∆b)
see [2, 6, 7, 23]. The degrees b for which H˜i(∆b)6=0 are called i-Betti degrees.
The semigroup A is pointed, so we can partially order A with the relation
c≥d⇐⇒ there is e∈A such that c=d+e.
The minimal elements ofthe set{b: βi,b(R/IL)6=0} withrespectto ≥ arecalled
minimal i-Betti degrees.
3. Simple syzygies
Inthissectionwepresentthe theoryofsimple syzygiesofA-homogeneousideals
for a positive grading A. We will apply these results for lattice ideals.
Let I be an A-homogeneous ideal and (F,φ) be a minimal A-graded free reso-
lution of R/I, where
F = R·E .
i M ti
1≤t≤si
In particular E is the basis element of F ∼= R of A-degree 0. The elements of
10 0
Imφ =kerφ arecalledi-syzygies. InthesequelallsyzygiesareA-homogeneous.
i+1 i
We note that the zero syzygies of R/I are the elements of I. Let h be an A-
homogeneouselement ofF . We write h asa combinationofthe basiselements E
i ti
with nonzero coefficients:
a
(1) h= X (X catx t)Eti .
1≤t≤si cat6=0
SYZYGIES 3
Since h is A-homogeneous, degAxat +degA(Eti) = degA(h). When h ∈ kerφi we
define S(h), the syzygy support of h to be the set
a
S(h)={x tEti : cat 6=0},
and partially order the A-homogeneous i-syzygies by h′ ≤h if and only if S(h′)⊂
S(h). WenotethatE mayappearinS(h)morethanoncewithdifferentmonomial
ti
coefficients.
Definition 3.1. We say that a nonzero A-homogeneous h ∈ kerφ is simple if
i
there is no nonzero A-homogeneous h′ ∈kerφ such that h′ <h.
i
We note that the syzygy support of h∈kerφ and thus also the simplicity of h
i
depends on the basis {E } of F . We also note that when I is a lattice ideal then
ti i L
h∈I is simple (as a zero syzygy of R/I ) if and only if h is a binomial. Next we
L L
define m-supp(h), the monomial support of h∈F . Let C ,C two fibers, T ⊂C ,
i 1 2 1 1
T ⊂ C . We let T ·T = {m m : m ∈ T }. Let h be a sum as in (1). We
2 2 1 2 1 2 i i
recursively define m-supp(h) by setting m-supp(E )={1} and
10
a
m-supp(h)= {x t}·m-supp(φ (E )).
[ i ti
xatEti∈S(h)
We note that if the A-degree of h is b then m-supp(h) ⊂ Cb. We also note that
m-supp(E )=m-supp(φ (E )).
ti i ti
If T is a subset of monomials then we set
gcd(T):=gcd(m: m∈T) .
Definition 3.2. For a vector b ∈ A we define the gcd-complex ∆ (b) to be the
gcd
simplicial complex with vertices the elements of the fiber deg−1(b) and faces all
L
subsets T ⊂deg−1(b) such that gcd(T)6=1.
L
Let b∈A. We remarkthat the gcd complex ∆gcd(b) and the complex ∆b have
the same homology, see [11].
Lemma 3.3. Let I be an A-homogeneous ideal, (F,φ) a minimal A-graded free
resolution of R/I and h a simple i-syzygy of A-degree b, i ≥ 1. Then m-supp(h)
is a connected subset of ∆ (b).
gcd
Proof. Let h be given as in equation (1). Since F is minimal it follows that
L
xat 6= 1, for all xat such that xatEti ∈ S(h), see [23]. Therefore for any fixed
xat, the set {xat}·m-supp(φi(Eti)) is a face of ∆gcd(b). Moreover, since i ≥ 1
and F is minimal, 1 ∈/ m-supp(φ (E )). Therefore for each E , the monomials
L i ti ti
in the sets {xat}·m-supp(φi(Eti)) where xatEti ∈S(h) are in the same connected
componentof∆ (b). Suppose now thatm-supp(h)is disconnectedandhas l≥2
gcd
components. This means that there are disjoint index sets J , r = 1,...,l such
r
that for each r
a
{x j}·m-supp(φ (E ))
[ [ i ji
j∈Jr xajEji∈S(h)
is a connected component of m-supp(h). Note that the different components of
m-supp(h) in ∆ (b) have no variable in common. It follows that for a fixed r
gcd
h′ = X(X cajxaj)Eji
j∈Jr caj6=0
is a syzygy and h′ <h, a contradiction. (cid:3)
4 H.CHARALAMBOUSANDA.THOMA
We also remark that if (F,φ) is a minimal A-graded free resolution of R/I and
{H :t = 1,...,b } is an A-homogeneous basis for F then by induction it can be
ti i i
shown that gcd(m-supp(H )) = 1. Next we will consider minimal free A-graded
ti
resolutions of R/I with a special property:
Theorem 3.4. There exists a minimal A-homogeneous generating set of kerφ
i
consisting of simple i-syzygies.
Proof. Since(F ,φ)isaminimalA-gradedfreeresolutionofS/I andAispointed
L L
kerφ canbegeneratedbyA-homogeneouselements. Itisenoughtoshowthatany
i
A-homogeneous i-syzygy can be written as a sum of simple A-homogeneous i-
syzygies. Leth∈kerφ notsimple. Wewilluseinductionon|S(h)|. Byhypothesis
i
there exists a simple h′ such that h′ <h. Since h′ 6=0 there is an a such that
h′ =ca′xaEa+ cb′xbEb, h=caxaEa+ cbxbEb, caca′ 6=0.
X X
b6=a b6=a
Note that if cb′ 6=0 then cb 6=0. It follows that
h=(h− ca h′)+ ca h′
ca′ ca′
while |S(h− ca h′)|<|S(h)|. Induction finishes the proof. (cid:3)
ca′
Definition 3.5. Let I be an A-homogeneous ideal, (F,φ) be an A-graded free reso-
lution of R/I and for each F we let B ={E } be an A-homogeneous basis for F .
i i ti i
We say that F is simple (with respect to the bases B ) if φ (E ) is simple for each
i i ti
i and t.
The corollary follows immediately from Theorem 3.4.
Corollary 3.6. Let I be an A-homogeneous ideal. There exists a minimal simple
A-graded free resolution of R/I.
We note that there might be more than one minimal simple free resolution of
R/I.
Example 3.7. Let I = hf ,...,f i be a complete intersection lattice ideal where
1 s
f :i=1,...,sisanR-sequenceofbinomialsandlet(K ,θ)betheKoszulcomplex
i •
onthe f . TheA-homogeneousstandardbasisofK consistsofelementsE where
i • J
J ranges over all subsets of [s]; if J ={k ,...,k }, where k <...<k then
1 t 1 t
E =e ∧···∧e , and θ(E )= (−1)i+1f E .
J k1 kt J X ki J\{ki}
(K ,θ) is a simple resolution of R/I (with respect to the basis {E }). Indeed let
• J
h 6= 0 be a syzygy such that h < θ(E ). Clearly this can only happen if |J| > 1.
J
Suppose that m E ∈/ S(h) while m E ∈S(h); where f =m −n ,f =
i J\{ki} j J\{kj} i i i j
m −n . Note that since R has no zero divisors and the coefficient of E
j j J\{ki,kj}
in θ(h) has to be zero, it follows that n E ∈ S(h). Therefore the coefficient
i J\{ki}
of E in θ(h) is (c n f −(c m −d n )f ) for some c ,c ,d ∈ k which
J\{ki,kj} i i kj j j j j ki i j j
cannot be zero since f ,f is an R-sequence, a contradiction.
ki kj
We note that a simple syzygy is determined by its syzygy support:
Theorem 3.8. If h and h′ are two simple i-syzygies and S(h)=S(h′), then there
exists a c6=0 in k such that h=ch′.
SYZYGIES 5
Proof. Since h′ 6= 0 there exists ca′ 6= 0 such that h′t = ca′xaEa+Pb6=acbxbEb.
Therefore ca 6=0 and
h′′ =h− ca h′ ∈kerφ ,
ca′ i
while S(h′′) S(h). Therefore h′′ =0 and h=ca/ca′h′. (cid:3)
Next we give the definition of an indispensable complex. We note the related
notion of rigidity, see [14].
Definition 3.9. Let I be an A-homogeneous ideal. We say that an A-graded com-
plex (G,θ) is an indispensable complex for R/I if for any simple minimal A-graded
free resolution (F,φ) of R/I (with respect to bases B of F ), there is an inclusion
j j
map i:(G,θ)−→(F,φ) so that the image of G is a subcomplex of F. In particular
for each j there is a subset B ′ of B so that the set i−1(B ′) is an A-homogeneous
j j j
basis of G . If H ∈i−1(B ′) we call θ (H ) an indispensable j−1 syzygy of R/I.
j tj j j tj
We say that G is a strongly indispensable complex for R/I if the above holds
without the requirement for (F,φ) to be simple.
Let I be a toric ideal. A polynomial of I is simple, as a zero syzygy, if and
L L
onlyifitisabinomial. Inthiscasetheindispensablezero-syzygiesofR/I arealso
L
strongly indispensable. More precisely in [10] it was shown that the binomial f of
degreebiscontainedineveryminimalsetofbinomialgeneratorsofI ifandonlyif
L
bis a minimal 1-Bettidegreeandthe fiber Cb consists ofjust two monomialswith
no common divisor: the difference of these two monomials is f up to a constant
multiple. Such a binomial is not only contained in every minimal set of binomial
generators of I ; it is necessarily contained in every minimal system of generators
L
of I and thus is strongly indispensable, see [20]. In [20], Peeva and Sturmfels
L
introduced the algebraic Scarf complex for R/I . They showed that this complex
L
iscontainedineveryminimalresolutionofR/I . ItfollowsthatthealgebraicScarf
L
complex is a strongly indispensable complex. We will generalize this construction
and show that the generalized algebraic Scarf complex is also indispensable.
4. The generalized Scarf complex
Let max (∅)=(−∞,...,−∞). For J ⊂L, 0<|J|<∞, we let
max (J)=(max{a : a∈J},··· ,max{a : a∈J})∈Zn .
1 n
Wehavemax (J′)≤max (J)ifandonlyifmax (J′) ≤max (J) whilemax (∅)
i i
isthe smallestelement. ForJ ⊂K ⊂Lwedefinev-supp (J), thevariable support
K
of J in K: if J =∅ we set v-supp (∅)=∅ and for all other J we set
K
v-supp (J):={i: ∃ a∈J, such that max (K) −a >0}.
K i i
We note that
v-supp (J)= supp(xmax (K)−a),
K [
a∈J
where supp(xa)={i:x |xa}.
i
In [20] the complex ∆ was defined to be the collection of all finite subsets J of
L
L with unique max (J). We extend this complex to ∆ as follows:
L
e
Definition 4.1. Let ∆ be the collection of all finite subsets J of L that satisfy
L
the following conditiones:
(1) if J′ (J then max (J′)<max (J)
6 H.CHARALAMBOUSANDA.THOMA
(2) if a∈/ J and |J|≤2 then a(cid:2)max (J)
(3) if a∈/ J, |J|>2 and a≤max (J) then
supp(xmax (J)−a) v-supp (J)=∅ .
\ J
We note that max (J) is determined by at most n elements of J. If J ∈∆ then
L
the first condition implies that |J| ≤ n. On the other hand if |J| = 2 tehen the
first two conditions imply that max (J) is unique. Finally all sets J with unique
max (J) are in ∆ and thus ∆ ⊂∆ .
L L L
e e
Proposition 4.2. ∆ is a simplicial complex.
L
e
Proof. We have {a} ∈ ∆ for every a ∈ L: b ≤ a =⇒ a − b ≥ 0 ∈ L, a
L
contradiction. The case |eJ| = 2 is trivial. We now examine the case |J| > 2, J ∈
∆ .
L
e Let a ∈ J; we will show that J = J \ {a} ∈ ∆ . Let J ( J such that
1 L 2 1
max (J2)=max (J1). Itfollowsthatmax (J2∪{a})e=max (J),acontradiction.
Suppose now that c ∈/ J and c ≤ max (J ). Therefore c 6= a, c < max (J)
1 1
and supp(xmax (J)−c) v-supp (J) = ∅. On the other hand since max (J ) <
T J 1
max (J), it follows that for some i, max (J ) < max (J) . Therefore a =
1 i i i
max (J) while for all b∈J , b <max (J) . In particular i∈v-supp (J). This
i 1 i i J
implies that i ∈/ supp(xmax (J)−c), and c = max(J) . Thus c (cid:2) max (J ), a
i i 1
contradiction. It follows that max (J ) is unique and J ∈∆ . (cid:3)
1 1 L
e
There is a natural action of the lattice L in ∆ , since J ∈ ∆ if and only if
L L
J +a ∈ ∆ for any a ∈ L. We identify ∆ weith its poset ofenonempty faces,
L L
and we foerm the quotient poset ∆ /L. Thies poset is called the generalized Scarf
L
complex of L. e
Proposition 4.3. The generalized Scarf complex ∆ /L is a finite poset.
L
e
Proof. Let ∆0 be the link to zero: ∆0 ={J ⊂L\{0}:J∪{0}∈∆ }. As in [20],
L L L
since L actsetransitively on the veretices of ∆ it is enough to shoew that ∆0 has
L L
finitely many vertices. The vertices a of ∆0eare such that {a,0} ∈ ∆ , theerefore
L L
max({a,0})=a+ is unique and we are exeactly in the case of [20, Proeposition2.2]:
therearefinitelymanyprimitiveelementsofLandaisoneofthem,seealso[3]. (cid:3)
For all J ⊂L we define
C :={xmax (J)−a : a∈J} .
J
We note that |C | = |J|. Moreover if xu ∈ C then deg xu = max (J)+L. It
J J L
followsthat C is a subset of the fiber deg−1(max (J)+L). We alsonote that C
J L J
is not necessarily an entire fiber and there are fibers or part of fibers that cannot
be expressed in the form C . The following Lemma determines exactly the cases
J
when this can happen.
Lemma 4.4. Let G be a subset of a fiber Cb. Then G = CJ for some J ⊂ L if
and only if gcd(G)=1.
Proof. Suppose that G = C for some J ⊂ L. For each 1 ≤ i ≤ n there exists
J
a ∈ J such that max (J) = a . It follows that i ∈/ supp(xmax (J)−a). Therefore
i i
gcd(G)=1.
SYZYGIES 7
Suppose that gcd(G) = 1 and xe ∈ G. Let J = {a ∈ Zn : xe−a ∈ G}.
Since deg (xe) = deg (xe−a) = b it follows that J is a subset of L. We claim
L L
that max (J) = e. Indeed, since xe−a ∈ G it is clear that e ≥ a and therefore
max (J)≥e. Moreovergcd(G)=1implies that foreach1≤i≤n there exists an
a∈J such that a =e . Therefore max (J)=e and G=C . (cid:3)
i i J
We isolate a slight variation of a useful remark of [20].
Lemma 4.5. If b is an i-Betti degree then Cb =CJ for some subset J of L.
Proof. Since H˜i(∆b) 6= 0 it is enough to show that gcd(Cb) = 1, see Lemma 4.4.
Indeed if gcd(Cb) 6= 1 then ∆b would be a cone with apex any variable in the
support of gcd(Cb) and ∆b would have no homology, a contradiction. (cid:3)
We will also make use of the following lemma:
Lemma 4.6. C =C if and only if J′ =u+J, for some u∈L.
J J′
Proof. One direction is direct. For the opposite let a ∈ J and assume that C =
J
C . Then there exists a′ ∈ J′ such that xmax (J)−a = xmax (J′)−a′ and thus
J′
max (J)−a=max (J′)−a′. WewillshowthatJ =J′−a′+a. Letb∈J′−a′+a.
Then b+a′−a∈J′. We note that CJ′−a′+a =CJ′ =CJ. Therefore
xmax (J′)−b−a′+a =x(max (J′)−a′)−b+a =xmax (J)−b ∈C
J
and b∈J. (cid:3)
When a fiber Cb is not of the form CJ some of its subsets may be expressed as
such.
Definition4.7. LetCb beafiberandG⊂Cb. WesaythatGisabasiccomponent
of Cb if the following are satisfied:
• G=C for some J ⊂L
J
• gcd(G\{m})6=1 for all m∈G and
• G is a connected component of ∆gcd(b) if |Cb|>2.
If Cb satisfies the above properties then we call Cb a basic fiber.
We note that not all fibers contain subsets that are basic fiber components. As a
matteroffactwewillshow,seeTheorem4.12,thatthesetofbasicfibercomponents
is finite. The definition of a basic fiber first appeared in [20]. It follows directly
from definition 4.7 that if Cb is a basic fiber and |Cb| > 2 then Cb has only one
connected component. Moreover if |Cb| = 2 then Cb is a basic fiber if and only if
Cb isdisconnected. WealsonotethatifGisabasiccomponentofCb and|Cb|=2
thenG=Cb isabasicfiber. ThisistheonlywayC canbeabasicfibercomponent
when |C|=2 as the following Lemma shows:
Lemma 4.8. If |J| = 2 and CJ is a basic component of Cb then CJ is a basic
fiber.
Proof. We have|C |=2. ByLemma4.4C isdisconnectedin∆ (b). Thethird
J J gcd
condition of definition 4.7 implies that CJ =Cb. (cid:3)
In [20, Theorem 3.2, Lemma 3.3] the following was shown:
Lemma 4.9. If Cb is a basic fiber with i+1 elements then dimH˜i(∆b)=1 and b
is a minimal i-Betti degree.
8 H.CHARALAMBOUSANDA.THOMA
Let T be a subset of a fiber Cb. We denote by [T] the set of monomials in T
divided by gcd(T). Let I′ =I∪{m}⊂T, m∈/ I. It is clear that
m
[T \I′]=[[T \I]\{ } ].
gcd(T \I)
Lemma 4.10. If CJ is a basic component of Cb and ∅ =6 I ⊂ CJ then [CJ \I] is
a basic fiber.
Proof. By the preceding comment it suffices to prove the statement when |I| = 1.
Let d = deg (gcd(C \I)) and b′ = b−d. Since gcd(C \I) 6= 1 it follows that
A J J
b′ 6= b. We will show that [CJ \I] = Cb′. Indeed if m ∈ Cb′ then mxd ∈ Cb
and clearly mxd is in same connected component of Cb as CJ \I, thus mxd ∈CJ
and [CJ \I]=Cb′. It is immediate that gcd[CJ \I]=1. The remaining condition
follows as in [20, Lemma 2.4]. (cid:3)
The next lemma helps in computing gcd(C \{m}).
J
Lemma 4.11. Let CJ be a subset of Cb, a∈J and m=xmax (J)−a. Then
gcd(C \{m})=xmax (J)−max (J\{a}) .
J
Proof. AnarbitraryelementofC \{m}isoftheformxmax (J)−bwhereb∈J\{a}.
J
We note that b≤max (J \{a}). It follows that
xmax (J)−b =xmax (J)−max (J\{a})x(max (J\{a})−b) .
For each 1 ≤ i ≤ n there exist a b ∈ J \{a} such that b = (max (J \{a}) .
i i
Therefore gcd(C \{m})=xmax (J)−max (J\{a}). (cid:3)
J
The set of basic fiber components forms a poset by setting C ≤ C if and
J′ J
only if there exists a monomial xr such that xrC ⊂ C . We state and prove the
J′ J
analogue of [20, Theorem 2.5].
Theorem 4.12. The poset of basic fiber components is isomorphic to the general-
ized Scarf complex ∆ /L and is finite.
L
e
Proof. Let F be an element in ∆ /L. Choose a representative J of F. Let b =
L
max (J)+L. For any other repreesentativeJ′ of F, CJ =CJ′ ⊂Cb. We will show
that CJ is a basic component of Cb.
Let m ∈ C . We will show that gcd(C \ {m}) 6= 1. Suppose that m =
J J
xmax (J)−a, for a∈J. Since J ∈∆ , we have that max (J\{a})<max (J) and
L
xmax (J)−max (J\{a}) 6=1. The deesired inequality now follows by Lemma 4.11.
Let |J| = 2. Then J ∈ ∆L implies that max (J) is unique and CJ = Cb.
Suppose now that |C | = |Je| > 2. Since gcd(C \ {m}) 6= 1, it follows that
J J
C \{m} is a face of ∆ (b) and thus C is connected. To show that C is a
J gcd J J
connectedcomponentofCb weconsideranelementxu ∈Cb\CJ. Thenu+L=b
and therefore u=max (J)−a where a∈L and a∈/ J. Since a≤max (J) and
supp(xmax (J)−a) v-supp (J)=∅
\ J
we have that gcd(xu,m) = 1, ∀ m ∈ C . It follows that C is a connected com-
J J
ponent of ∆ (b). Set C = C . It is immediate that ψ : F 7→ C is order
gcd F J F
preserving.
Wewillshowthatψisbijective. TheinjectivityfollowsimmediatelyfromLemma
4.6. To show that ψ is surjective we let CJ be a basic component of Cb. We need
SYZYGIES 9
to show that J ∈ ∆ . Let J′ ( J. For I = {xmax (J)−a : a ∈/ J′} we have
L
that gcd(C \ I) =egcd({xmax (J)−a : a ∈ J′}) 6= 1. Therefore there is an i
J
such that max (J) > a for all a ∈ J′. It follows that a < max (J), ∀ a ∈ J′
i i
and max (J′) < max (J). Suppose now that a ∈ L is such that a ∈/ J and
a≤ max (J). It follows that max (J ∪{a})= max (J) and CJ ( CJ∪{a} ⊂ Cb.
If |J|=2 then Lemma 4.8 gives a contradiction, so in this case max (J) is unique
and J ∈∆L. Suppose now that |J|>2. Since m=xmax (J)−a ∈Cb\CJ and CJ
is a conneected component of ∆gcd(b), it follows that no variable in the support of
m is in the support of any monomial in C and thus
J
supp(xmax (J)−a) v-supp (J)=∅
\ J
as required. (cid:3)
5. The generalized algebraic Scarf complex
We generalize the notion of the algebraic Scarf complex introduced in [20].
Definition 5.1. The generalized algebraic Scarf complex is the complex of free
R-modules
(G ,θ) := R·E
L M C
e
C∈∆L/L
where E denotes a basis vector in homological degree |C|−1 and the sum runs
C
over all basic fiber components C, identified as elements of ∆ /L. We let
L
e
θ(E )= sign(m,C)gcd(C\{m})E ,
C X [C\{m}]
m∈C
where sign(m,C) is (−1)l+1 if m is in the lth position in the lexicographic ordering
of C.
Our firstremarkis that (G ,θ) is a subcomplex of the Taylorresolution,see [4,
L
Proposition3.10]. Indeed the canonicalbasis of the Taylorcomplex in homological
degree i consists of vectors E where C is a subset of a fiber such that gcd(C)=1
C
and|C|=i+1. Thedifferentialof(G ,θ)istherestrictionofthedifferentialofthe
L
Taylor complex on the elements of (G ,θ). Moreover we note that the algebraic
L
Scarfcomplex of [20] is a subcomplex of (G ,θ). Indeed the canonicalbasis of the
L
Scarf complex in homological degree i consists of vectors E where C is a basic
C
fiber and the differential coincides for these elements. We also note that for i ≤ 1
the algebraic Scarf complex is identical to the generalizedalgebraic Scarf complex.
Theorem 5.2. The complex G is an indispensable complex for R/I .
L L
Proof. Let (F ,φ) be a simple A-graded minimal resolution of R/I with respect
L L
toanA-homogeneousbasis{H , t=1,...,b }ofF . Wewilluseinductiononthe
tj j j
homological degree i, the case i = 0 being trivial. For i = 1, let C = {m ,m } be
1 2
a basic fiber. Then θ (E )=m −m is an indispensable binomial of I meaning
1 C 2 1 L
thatup to a constantmultiple, θ (E ) is partofanyminimal systemofgenerators
1 C
of the I , see [10], [20]. Note that binomials in a lattice ideal are always simple.
L
Thus for i≤1, θ (E ) are indispensable.
i C
Suppose now that G is indispensable for homological degrees less than i and
L
thus φj|Gj = θj for j < i. Let C be a basic component of Cb with cardinality
i+1. It is clear that θ (E ) ∈ kerφ . We will show that θ (E ) is a simple
i C i−1 i C
10 H.CHARALAMBOUSANDA.THOMA
syzygy. The proof is essentially the same as in Example 3.7. Indeed suppose that
0 6= h ∈ kerφ and h < θ (E ). Since S(h) ⊂ S(θ (E )) it follows that h ∈
i−1 i C i C
G and φ (h) = θ (h). Moreover since S(h) ( S(θ (E )) and S(θ (E )) =
i−1 i−1 i−1 i C i C
{ gcd(C\{m})E : m∈C}, it follows that for some monomial m ∈C,
[C\{m}] 1
gcd(C\{m })E ∈/ S(h).
1 [C\{m1}]
On the other hand since h6=0 we can find m ∈C such that
2
gcd(C\{m })E ∈S(h).
2 [C\{m2}]
It follows that gcd(C \{m ,m })E ∈ S(φ (h)), a contradiction since
1 2 [C\{m1,m2}] i−1
φ (h)=0.
i−1
Next suppose that θ (E ) is an R-linear combination of i-syzygies of strictly
i C
smaller A-degree. It follows that F in homological degree i has a basis generator
L
h of A-degree b1, where b1 <b and
m-supp(θi(EC))∩m-supp(mh)6=∅, for some m∈Cb−b1.
By Lemma 4.5 we have that gcd(Cb ) = 1. Therefore mCb ∩C 6= ∅. Since C
1 1
is a connected component it follows that mCb ⊂ C and therefore Cb = [C \I]
1 1
for some I ⊂ C. Lemma 4.10 implies that Cb is basic while Lemma 4.9 implies
1
that Cb has at least i +1 elements. Recall that by construction |C| = i+ 1.
1
Since gcd(C) = 1, mCJ1 ⊂ C and |mCb1| = i+1 it follows that |C| > i+1, a
contradiction.
WestillneedtoshowthatwecanidentifyE withaconstantmultipleofabasis
C
element H for some t. Suppose that θ (E )= c φ (H ) where deg (H )=b
ti i C t i ti A ti
P
for at least one t. Then for some t with deg (H )=b, we have that
A ti
gcd(C\{m})E ∈S(θ (E ))∩S(φ (H )) ,
[C\{m}] i C i ti
for a monomial m∈C. Therefore m-supp(θ (E ))∩m-supp(φ (H ))6=∅.
i C i ti
Since m-supp(θ (E )) = C, m-supp(H ) = m-supp(φ (H )) and by Lemma 3.3
i C ti i ti
m-supp(φ (H )) is connected, it follows that m-supp(H )⊂C. Suppose that
i ti ti
φ (H )= p H
i ti X s s,i−1
s
where Hs,i−1 are basis generators of Fi−1 of A-degree bs and ps ∈ R is A-homo-
geneous of A-degree b−bs. It is clear that bs < b and that Cb < C. Moreover
s
H˜i−1(∆bs)6=0andthereforeCbs =[C\I]forsomesubsetI ofC. Wenotethatthe
set I need not be unique. By Lemma 4.10 it follows that Cb is a basic fiber. By
s
inductionE isindispensableandistheuniquebasiselementofF ofA-degree
Cbs i−1
bs, up to a constant multiple. By Lemma 4.9 |Cb | ≥ i. Since Cb = [C \I], it
s s
followsthat|Cbs|≤i. Therefore|Cbs|=iandCbs =[C\{ml}]foramonomialml.
Let cxγ be a monomial term of p where c∈k−{0}. Since m-supp(φ (H ))⊂C,
s i ti
gcd(C) = 1 and gcd(Cbs) = 1 it follows that xγ = gcd(C \{ml}), so actually
p = cxγ for a c 6= 0. Thus φ (H ) ≤ θ (E ) and since both are simple we
s i ti i C
get that S(φ (H )) = S(θ (E )). We apply Theorem 3.8 to obtain the desired
i ti i C
conclusion. (cid:3)
Next we consider a complex that sits between the algebraic Scarf complex and
the generalized algebraic Scarf complex.
