Minimal free resolutions of lattice ideals of digraphs
Liam O'Carroll, Francesc Planas-Vilanova
TL;DR
This paper constructs minimal free resolutions for lattice ideals derived from finite, strongly connected, weighted directed graphs, extending previous work on undirected graphs and providing explicit algorithms for syzygy computation.
Contribution
It introduces a method to compute minimal free resolutions of lattice ideals for directed graphs, refining existing algorithms and characterizing minimality in strongly complete digraphs.
Findings
Provides a free resolution for lattice ideals of directed graphs.
Characterizes when the resolution is minimal (strongly complete digraphs).
Extends undirected graph results to directed, weighted graphs.
Abstract
Based upon a previous work of Manjunath and Sturmfels for a finite, complete, undirected graph, and a refined algorithm by Er\"ocal, Motsak, Schreyer and Steenpa{\ss} for computing syzygies, we display a free resolution of the lattice ideal associated to a finite, strongly connected, weighted, directed graph. Moreover, the resolution is minimal precisely when the digraph is strongly complete.
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Taxonomy
TopicsTopological and Geometric Data Analysis · Commutative Algebra and Its Applications · Advanced Combinatorial Mathematics
Minimal free resolutions of lattice ideals of digraphs
Liam O’Carroll and Francesc Planas-Vilanova
Abstract.
Based upon a previous work of Manjunath and Sturmfels for a finite, complete, undirected graph, and a refined algorithm by Eröcal, Motsak, Schreyer and Steenpaß for computing syzygies, we display a free resolution of the lattice ideal associated to a finite, strongly connected, weighted, directed graph. Moreover, the resolution is minimal precisely when the digraph is strongly complete.
2010 Mathematics Subject Classification:
13D02,13P10,05C25,05C50,05EXX
1. Introduction
The aim of this paper is to present a free resolution of the lattice ideal I(\mbox{\mathcal{L}}) associated to the lattice spanned by the columns of the Laplacian matrix of a finite, strongly connected, weighted, directed graph , and to show that this resolution is minimal if and only if the graph is strongly complete. Our techniques are a novel mixture of a mild generalization of previous work by Manjunath and Sturmfels for finite, complete, weighted, undirected graphs and a recent refined algorithm for computing syzygies due to Eröcal, Motsak, Schreyer and Steenpaß. (The meaning of the terminology we use is presented in Section 2.)
First we present some background.
The Abelian Sandpile Model (ASM) is a game played on a finite, weighted, connected, undirected graph with vertices, that realizes the dynamics implicit in the discrete Laplacian matrix of the graph, this matrix being an integer matrix that is symmetric. Each configuration is a mapping from the vertices of the graph into the set of non-negative integers. The value of the mapping at a vertex may be considered as the number of grains of sand on a sandpile placed at the vertex. The game’s evolution is given by a ‘toppling’ rule: each vertex containing at least as many grains as it has neighbours distributes one grain to each of them. (This process has also been called ‘sand-firing’ or ‘chip-firing’.)
The ASM was introduced by Bak, Tang and Wiesenfeld [BTW] in the context of self-organized critical phenomena in statistical physics and has been studied extensively since. In the seminal paper [CRS], Cori, Rossin and Salvy, enumerated the vertices of the undirected graph using a natural metric, considered the lattice spanned over by the rows of the symmetric Laplacian matrix and introduced a so-called toppling (or lattice) ideal I(\mbox{\mathcal{L}}) in the polynomial ring \mbox{\mathbb{Q}}[x]:=\mbox{\mathbb{Q}}[x_{1},\ldots,x_{n}]. This ideal encodes configurations with monomials and topplings with binomials. In particular, they showed that the set of toppling binomials constituted a Gröbner basis for I(\mbox{\mathcal{L}}) with respect to a natural monomial ordering in \mbox{\mathbb{Q}}[x], which may be taken to be the graded reverse lexicographic ordering.
In their recent survey paper [PPW], Perkinson, Perlman and Wilmes associated to the Laplacian matrix of a directed graph a collection of binomial equations, and noted that these binomials span the lattice ideal I(\mbox{\mathcal{L}}) in \mbox{\mathbb{C}}[x_{1},\ldots,x_{n}], where again is the lattice spanned over by the rows of the (now non-necessarily symmetric) Laplacian matrix , using the theory of lattice ideals extensively in their discussion.
More recently, and for complete undirected graphs, i.e., any two distinct vertices are connected by an edge, Manjunath and Sturmfels [MaS] gave a minimal free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}), an arbitrary field, showing also that the binomials in question formed a minimal generating set and a minimal Gröbner basis of I(\mbox{\mathcal{L}}) in the graded reverse lexicographic ordering. Subsequently, Manjunath, Schreyer and Wilmes [MSW], and Mohammadi and Shokrieh [MoS], gave a minimal free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) in the case where the underlying undirected graph was, more generally, connected. We remark here that the techniques used in these papers are a mix of results from combinatorial graph theory and the theory of Gröbner bases (e.g., the Schreyer algorithm as a tool to proving the exactness of a complex).
The transition from undirected to directed graphs introduces extra technicalities, and means that we have to rely heavily on techniques from the theory of Gröbner bases and homogeneous Commutative Algebra. Our general setting is the following: is a finite, strongly connected, weighted, directed graph, stands for its Laplacian matrix, is the lattice spanned by the columns of and I(\mbox{\mathcal{L}}) is the lattice ideal associated to .
First of all, we see that, having again enumerated the vertices of the graph using a generalized natural metric, I(\mbox{\mathcal{L}}) is homogeneous in a weighted reverse lexicographic order on \mbox{\mathbb{K}}[x], an arbitrary field, the weighting coming from the structure of \mbox{{\rm adj}}(L) (see Remark 2.8). We then use basic aspects of homogeneous Commutative Algebra and previous work of ours to show that I(\mbox{\mathcal{L}}) is a Cohen-Macaulay ideal of dimension , so that \mbox{{\rm projdim}}_{\mathbb{K}[x]}(\mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}))=n-1: see Corollary 2.10.
This earlier work shows up as follows. We find (see Sections 2.2, 2.3) that the matrices we deal with are what we called Critical Binomial Matrices ( matrices, in brief) in previous work, and that is, further, strongly complete precisely when the matrices are what we previously called Positive Critical Binomial Matrices ( matrices, in brief): see [OP], [OPV]. In fact, as we shall see in Section 2, since we treat the case where is strongly connected, the matrix has a further property which is precisely Irreducibility, hence the matrices are Irreducible Critical Binomial Matrices ( matrices, in brief), whose properties are developed in the paper: see, for example, Sections 2.6, 2.7, 4.1. Wherever possible, we give new proofs and develop new approaches to this previous work: see, for example, Propositions 2.3, 2.9.
There are two main ingredients in the proof of our main theorem (Theorem 6.1). The first is the fact that the resolving complex \mbox{{\rm Cyc}}_{n} of [MaS] (treated with a little care in our more general context, as the Laplacian matrix need no longer be symmetric) continues to be a complex and, moreover, is exact in degree zero: see Sections 3.5, 4.5. The second ingredient is a refined choice of Schreyer syzygies: see Section 5.2. We find, via an inductive proof that keeps track of the leading terms of the Manjunath-Sturmfels boundaries, that in a very satisfying yet surprising manner, these boundaries pair off with appropriate Schreyer syzygies, enabling one to deduce in Theorem 6.1 that \mbox{{\rm Cyc}}_{n} is in fact acyclic. It then follows easily that \mbox{{\rm Cyc}}_{n} is minimal if and only if is, further, strongly complete (see Corollary 6.16).
We finish this introduction by emphasizing that, usually, and alternatively to the works of, e.g., Perkinson, Perlman and Wilmes in [PPW], or Backman and Manjunath in [BM], we have taken to be the lattice spanned by the columns of the Laplacian matrix and not the rows (see Sections 2.1 and 2.5). Note that, for the undirected graph case this is an irrelevant matter, because being a symmetric matrix, then the lattice spanned by the columns or the lattice spanned by the rows do coincide. It is worth to mention here that studying the case of the lattice spanned by the columns can be related to a column chip-firing game. This is done by Asadi and Backman in [AB, Introduction and Section 3.2]. An interesting future research would be to pass from columns to rows. As a possible approach one could consider the lattice \mbox{\mathcal{L}}^{\top}, spanned by the columns of the transpose of the Laplacian matrix , and then proceed in a similar vein as in [OPV]. See also the connection with Eulerian digraphs (the indegree and outdegree coincide for every vertex) in the aforementioned work of Asadi and Backman ([AB, Theorem 3.18]).
Throughout, we use the explicit case as a running illustrative example.
Liam O’Carroll died on October 2017, while this manuscript was under revision. Liam and I started to work together several years ago. To me, all these years of collaboration with him have been, above all, years of enjoying our friendship. I will greatly miss him.
2. Notations and Preliminaries
2.1. General setting
The following notations will remain in force throughout the paper. We denote the set of non-negative integers by and the set of positive integers by \mbox{\mathbb{N}}_{+}. Let be a positive integer with , to avoid trivialities, and let . Any vector \alpha\in\mbox{\mathbb{Z}}^{n} can be written uniquely as , where \alpha^{+}=\max(\alpha,0)\in\mbox{\mathbb{N}}^{n} and \alpha^{-}=-\min(\alpha,0)\in\mbox{\mathbb{N}}^{n} have disjoint support. Set \mbox{1\mkern-4.6mu{\rm l}}=(1,\ldots,1).
Let be a field and let \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x_{1},\ldots,x_{n}]=\mbox{\mathbb{K}}[x] be the polynomial ring over in indeterminates . The maximal ideal generated by will be denoted \mbox{\mathfrak{m}}=(x_{1},\ldots,x_{n}). For any \alpha=(\alpha_{1},\ldots,\alpha_{n})\in\mbox{\mathbb{N}}^{n}, we write as shorthand for . By a binomial in , we understand a polynomial of with at most two terms, say , where \lambda,\mu\in\mbox{\mathbb{K}} and \alpha,\beta\in\mbox{\mathbb{N}}^{n}. A binomial ideal of is an ideal generated by binomials.
Let be an integer matrix, ; that is, m_{i,j}\in\mbox{\mathbb{Z}}, for all , . We will say that is non-negative, and write , if all the entries in are non-negative, i.e., , for all . Similarly, is said to be positive, , if , for all .
We will denote by and the -th row and -th column, repectively, of . We will denote by \mbox{\mathcal{M}}\subset\mbox{\mathbb{Z}}^{n} the additive subgroup of \mbox{\mathbb{Z}}^{n} spanned by the columns of , commonly called the lattice of \mbox{\mathbb{Z}}^{n} defined by the columns of .
Furthermore, will denote the binomial of defined by the -th column of . The ideal , generated by the binomials , is called the binomial ideal associated to the matrix . The binomial ideal I(\mbox{\mathcal{M}})=(x^{m^{+}}-x^{m^{-}}\mid m\in\mbox{\mathcal{M}}) is called the lattice ideal associated to . Clearly I(M)\subseteq I(\mbox{\mathcal{M}}).
By a permutation of a square matrix , we understand a permutation of the rows of together the same permutation of the columns. Equivalently, a permutation of is a square matrix , where is a permutation matrix resulting from a permutation of the columns of the identity matrix .
2.2. (Positive) critical binomial matrices and ideals
Recall that a critical binomial matrix, matrix for short, is an integer matrix defined as follows:
[TABLE]
where , , , for all , and, for each column of , at least one off-diagonal entry is nonzero (see [OPV]). Alternatively, an integer matrix is a matrix whenever , where is a positive, diagonal integer matrix and is a non-negative integer matrix with zero entries in the diagonal, nonzero columns , and such that (D^{-1}B)\mbox{1\mkern-4.6mu{\rm l}}^{\top}=\mbox{1\mkern-4.6mu{\rm l}}^{\top}. In particular, is an stochastic matrix. The expression , which is clearly unique, will be called the “ form of ”.
Observe that a permutation of a matrix is again a matrix. Indeed, writing in its form, then , and is diagonal and , with zero entries in the diagonal, nonzero columns and (P^{\top}DP)^{-1}(P^{\top}BP)\mbox{1\mkern-4.6mu{\rm l}}^{\top}=\mbox{1\mkern-4.6mu{\rm l}}^{\top}.
Since and are nonzero, for each column of an matrix , the binomial
[TABLE]
has two non-constant terms. The binomial ideal , generated by the , is called the critical binomial ideal, ideal for short, associated to (see [OPV]).
A special class of matrices is the following. When all the coefficients are positive, the matrix is said to be a positive critical binomial matrix ( matrix, for short). The ideal is called the positive critical binomial ideal ( ideal, for short) associated to . These ideals were extensively studied in [OP] (see also [OPV]).
2.3. Irreducible binomial matrices
A square matrix is called reducible if the set of indices can be split into two non-empty disjoint sets such that , for all and . The matrix is called irreducible if it is not reducible. Equivalently, is reducible if and only if there exists a permutation matrix such that
[TABLE]
where and are square matrices (see, e.g., [G, Chapter XIII, p.50]; see also [BR, § 3.2]). Here we understand as the submatrix of the matrix defined by the set of rows and the set of columns . Of course, this is equivalent to saying that there exists a permutation matrix that permutes columns and then corresponding rows, such that
[TABLE]
Clearly, adding a diagonal matrix, multiplying by a diagonal matrix, or transposing, are three operations that preserve the condition of irreducibility on a matrix.
We will say that is an irreducible critical binomial matrix ( matrix, for short), if is a matrix which is irreducible. Write in its form. If is irreducible, then is also irreducible, and conversely.
Clearly, we have the chain of implications linking these three properties: \mbox{\mathrm{PCB}}\Rightarrow\mbox{\mathrm{ICB}}\Rightarrow\mbox{\mathrm{CB}}. (When , the three classes coincide.) If , then the condition is equivalent to the condition , but there are matrices which are not . If , it is easy to see that the three classes are distinct.
Remark 2.1**.**
Let be a reducible matrix, . After a permutation, we can suppose that can be written as
[TABLE]
where is a square integer matrix, is a square integer matrix, with and . Note that L_{I,I}\mbox{1\mkern-4.6mu{\rm l}}^{\top}=0. In particular, since the sum of the entries of the first row is zero and , then . Similarly, , otherwise the last column of would not have an off-diagonal nonzero entry.
2.4. The digraphs we will deal with
All graphs \mbox{\mathbb{G}}=(\mbox{\mathbb{V}},\mbox{\mathbb{E}},w) we will deal with are assumed to be finite, weighted, directed graphs. That is, \mbox{\mathbb{V}}=\{v_{1},\ldots,v_{n}\} is the finite set of vertices; is the set of directed arcs of , which are ordered pairs , with ; and is a weight function that associates to a every arc in a weight w_{e}\in\mbox{\mathbb{N}}_{+}. Moreover, we will always assume that has no loops, sources or sinks. That is, there are no arcs from to , for any ; and for every vertex , there exists at least a and an arc (v_{j},v_{i})\in\mbox{\mathbb{E}}, and there exists at least a and an arc (v_{i},v_{k})\in\mbox{\mathbb{E}}. (See, e.g., [BR] or [GR], for all unexplained notations on graph theory.)
Assumption 2.2**.**
Thus, from now on, a digraph will mean a finite, weighted, directed graph, without loops, sources or sinks.
A digraph is said to be strongly connected if any two (distinct) vertices can be joined by a (directed) path. A directed path from vertex to a distinct vertex will be denoted by . The unweighted length of the path is the number of arcs which constitute it, without reference to their weights. The unweighted distance \mbox{{\rm d}^{\vee}}(v_{i},v_{j}) between distinct vertices and is defined to be the minimum unweighted length of a directed path connecting with . Thus, \mbox{{\rm d}^{\vee}}(v_{i},v_{j})=1 if and only if (v_{i},v_{j})\in\mbox{\mathbb{E}}. It may happen that \mbox{{\rm d}^{\vee}}(v_{i},v_{j})\neq\mbox{{\rm d}^{\vee}}(v_{j},v_{i}).
A directed graph is said to be strongly complete if any two (distinct) vertices can be joined by an arc, that is, (v_{i},v_{j})\in\mbox{\mathbb{E}}, for all , .
The adjacency matrix A(\mbox{\mathbb{G}}) of is the integer matrix given by A(\mbox{\mathbb{G}})_{i,j}=w_{e} if e=(v_{i},v_{j})\in\mbox{\mathbb{E}}, and A(\mbox{\mathbb{G}})_{i,j}=0 otherwise. Since has no loops, A(\mbox{\mathbb{G}})_{i,i}=0; since has no sources, for each column of A(\mbox{\mathbb{G}}), at least one off-diagonal entry is nonzero; since has no sinks, for each row of A(\mbox{\mathbb{G}}), at least one off-diagonal entry is nonzero. The (out)degree matrix D(\mbox{\mathbb{G}}) of is the integer diagonal matrix defined by D(\mbox{\mathbb{G}})_{i,i}=\sum_{e\in O(v_{i})}w_{e}, the weighted (out)degree of the vertex , where is the set of vertices , , such that (v_{i},v_{j})\in\mbox{\mathbb{E}}. Since has no sinks, D(\mbox{\mathbb{G}})_{i,i}\neq 0 for every . The Laplacian matrix L(\mbox{\mathbb{G}}) of is defined as L(\mbox{\mathbb{G}})=D(\mbox{\mathbb{G}})-A(\mbox{\mathbb{G}}).
2.5. Our dictionary: CB matrices - digraphs
Clearly, the Laplacian matrix L(\mbox{\mathbb{G}}) of a digraph is a matrix, sometimes also denoted (recall that we are using Assumption 2.2). Conversely, given an matrix , one can define a digraph \mbox{\mathbb{G}}_{L} of vertices \mbox{\mathbb{V}}=\{v_{1},\ldots,v_{n}\} with adjacency matrix in an obvious way, namely, there exists a directed edge between distinct vertices and if and only if , in which case the edge is assigned weight .
Moreover, is an matrix if and only if \mbox{\mathbb{G}}_{L} is strongly connected (see [CV], for a recent related work). Indeed, first remark that a permutation of is just a re-enumeration of the vertex set of and a corresponding relabelling of the directed arcs. Now suppose that is reducible and let be the partition on such that , for all and all . Then it follows that there is no directed path from a vertex to a vertex , with and . Conversely, suppose that is not strongly connected and, after a re-enumeration of the vertices and corresponding relabelling of the directed arcs, suppose there is no directed path from to . Let and let . After a re-enumeration of the vertices and corresponding relabelling of the directed arcs, we can suppose that and . Clearly, there is not an arc from , , to , (otherwise there would be a directed path and would belong to , a contradiction). Therefore and is reducible (see, e.g., [BR, Theorem 3.2.1], for more details).
Finally, is a matrix if and only if \mbox{\mathbb{G}}_{L} is strongly complete.
Observe that if is undirected, then is symmetric, and the discussion above can be specialized to this case in an obvious manner (see, e.g., [MaS]).
2.6. Irreducible critical binomial matrices and their adjugate
matrices
Given an integer matrix , let denote the matrix obtained from by eliminating the -th row and the -th column of . Set \mbox{{\rm adj}}(L)=((-1)^{i+j}\lvert L_{j,i}\rvert) to be the adjugate matrix of , sometimes also called the adjoint matrix, where stands for the determinant . For convenience, we write to denote .
In the next set of results, invertibility, ranks or linear subspaces are thought of over . Observe that if \mbox{{\rm rank}}(L)=n, then \mbox{{\rm adj}}(L) is invertible (since L\cdot\mbox{{\rm adj}}(L)=\rvert L\rvert\cdot\mbox{{\rm I}}). If \mbox{{\rm rank}}(L)=n-1, then, by definition, at least one minor of is nonzero, thus \mbox{{\rm adj}}(L)\neq 0; moreover, \mbox{{\rm rank}}(\mbox{{\rm adj}}(L))=1 because L\cdot\mbox{{\rm adj}}(L)=\lvert L\rvert\cdot\mbox{{\rm I}}=0 and so all the columns of \mbox{{\rm adj}}(L) belong to the nullspace of , which is of dimension . If \mbox{{\rm rank}}(L)\leq n-2, then all the minors of are zero, so \mbox{{\rm adj}}(L)=0. In particular, is invertible if and only if \mbox{{\rm adj}}(L) is invertible.
The next result can be deduced from [OPV, Theorem 7.6]. We present here a more direct proof, that does not involve a wider class of matrices.
Proposition 2.3**.**
Let be an integer matrix and let \mbox{{\rm adj}}(L) be its adjugate matrix.
If L\mbox{1\mkern-4.6mu{\rm l}}^{\top}=0, then all the rows of \mbox{{\rm adj}}(L) are equal.
If is a matrix, then, further, \mbox{{\rm adj}}(L)\geq 0.
If is an matrix, then, further, \mbox{{\rm rank}}(L)=n-1 and \mbox{{\rm adj}}(L)>0.
Proof.
If \mbox{{\rm rank}}(L)\leq n-2, then \mbox{{\rm adj}}(L)=0 and all the rows are equal. Suppose that \mbox{{\rm rank}}(L)\geq n-1. Since L\mbox{1\mkern-4.6mu{\rm l}}^{\top}=0, then \mbox{{\rm rank}}(L)=n-1 and so \mbox{{\rm rank}}(\mbox{{\rm adj}}(L))=1. Moreover, L\cdot\mbox{{\rm adj}}(L)=0, so every column of \mbox{{\rm adj}}(L) is in the -linear subspace \mbox{{\rm Nullspace}}(L), which is generated by \mbox{1\mkern-4.6mu{\rm l}}^{\top}. Hence there exist \mu_{1},\ldots,\mu_{n}\in\mbox{\mathbb{Q}} such that the columns of \mbox{{\rm adj}}(L) are equal to \mu_{1}\mbox{1\mkern-4.6mu{\rm l}}^{\top},\ldots,\mu_{n}\mbox{1\mkern-4.6mu{\rm l}}^{\top}:
[TABLE]
Thus each row of \mbox{{\rm adj}}(L) is equal to , which proves . Note that, since \mbox{{\rm adj}}(L) is an integer matrix, then necessarily \mu_{1},\ldots,\mu_{n}\in\mbox{\mathbb{Z}}.
The proof of is an adaptation of [OP, Lemma 2.1] (see also [OPV, Theorem 6.3, ]). Fix . Let us prove that is non-negative. By the Gershgorin Circle Theorem, every eigenvalue of lies within at least one of the discs \{z\in\mbox{\mathbb{C}}\mid\lvert z-a_{j,j}\rvert\leq R_{j}\}, , where . Observe that due to the fact that is a matrix. If \lambda\in\mbox{\mathbb{R}}, then . If \lambda\not\in\mbox{\mathbb{R}}, then since is a real matrix, the conjugate must also be an eigenvalue of . Since is the product of the (possibly repeated) eigenvalues of , it follows that and so \mbox{{\rm adj}}(L)\geq 0.
We now turn to the proof of . Let us first see that \mbox{{\rm adj}}(L)\neq 0. Write , with diagonal and stochastic. In particular, the spectral radius of is ([G, Chapter XIII, p. 83]). Since is irreducible, is an irreducible, non-negative, integer matrix with . By the Perron-Frobenius Theorem (see, e.g., [G, Chapter XIII, p. 53], [GR, Theorem 8.8.1, p. 178]), is realized by a simple eigenvalue; in particular, \mbox{{\rm Nullspace}}((D^{-1}B)^{\top}-\mbox{{\rm I}}) has dimension . Since \mbox{{\rm Nullspace}}((D^{-1}B)^{\top}-\mbox{{\rm I}}) coincides with \mbox{{\rm Nullspace}}(L^{\top}), it follows that and have rank . In particular, \mbox{{\rm adj}}(L) has rank and \mbox{{\rm adj}}(L)\neq 0.
Let us now prove that \mbox{{\rm adj}}(L)>0. Suppose not, i.e., there exists some which is zero. Since \mbox{{\rm adj}}(L)\neq 0, by permuting , we can suppose that are positive and are zero, where . Setting and , can be written as
[TABLE]
and and are two square matrices. Write , with and . Since \mbox{{\rm adj}}(L)\cdot L=0, then . It follows that . But and , for all and all , so . Thus is reducible, a contradiction. ∎
Notation 2.4**.**
Given an matrix, we will denote the last (and so any) row of \mbox{{\rm adj}}(L) by \mu=\mu(L)=(\mu_{1},\ldots,\mu_{n})\in\mbox{\mathbb{N}}^{n}_{+}. If , set .
Corollary 2.5**.**
Let be an integer matrix. Then is an matrix if and only if is a matrix and any set of rows of is linearly independent.
Proof.
Let be the -linear subspace spanned by the rows of a matrix . Writing \mbox{{\rm adj}}(L) as in (21), and since \mbox{{\rm adj}}(L)\cdot L=0, it follows that , with each . Suppose first of all that is irreducible. Fix . By Proposition 2.3, \mbox{{\rm rank}}(L)=n-1 and , for all . Therefore, has dimension and equals . Hence the set of rows is linearly independent. Conversely, if , the condition implies the condition . Suppose and that is reducible. After a permutation, one can suppose that can be written as in (12), where is an matrix, , such that L_{I,I}\mbox{1\mkern-4.6mu{\rm l}}^{\top}=0. In particular, \mbox{{\rm rank}}(L_{I,I}^{\top})=\mbox{{\rm rank}}(L_{I,I})\leq r-1. Therefore the first rows of are linearly dependent. ∎
Remark 2.6**.**
Clearly, an easy adaptation of the same argument shows that if is an matrix, then is a matrix and any set of columns of is linearly independent. However, the converse is not true, as the example below shows:
[TABLE]
2.7. Homogeneity of lattice ideals of strongly connected digraphs
Assumption 2.7**.**
Let be an matrix (or, equivalently, let \mbox{\mathbb{G}}_{L} be a strongly connected digraph). Let be defined as is Notation 2.4. From now on, we will always consider the -grading on \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x_{1},\ldots,x_{n}] where each is given weight . Thus for a monomial x^{\alpha}\in\mbox{\mathbb{A}}, we define the degree of as follows: .
Remark 2.8**.**
With this grading, the matrix ideal and the lattice ideal I(\mbox{\mathcal{L}})=(x^{l^{+}}-x^{l^{-}}\mid l\in\mbox{\mathcal{L}}) are homogeneous ideals of (cf. Definitions in Section 2.1). Indeed, let l\in\mbox{\mathcal{L}}. So there exists b\in\mbox{\mathbb{Z}}^{n} with , a linear combination of the columns of . Then , so and . Thus is homogeneous. (See [OP, Definition 2.2] or [OPV, Remark 5.6].)
Note that the hypothesis that be strongly connected is essential to define a -grading on \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x_{1},\ldots,x_{n}] and to ensure that is homogeneous. See, for instance, the example in Remark 2.6.
The following result can be deduced from [OPV, Proposition 7.6]. Here we give a shorter direct proof.
Proposition 2.9**.**
Let be an matrix. Then \mbox{{\rm rad}}(I(L),x_{i})=\mbox{\mathfrak{m}}, for all .
Proof.
Fix . Let \mbox{\mathbb{G}}_{L} be the corresponding strongly connected digraph. Let be a directed path passing through all the vertices and begining in (there might be possibly repeated vertices, though adjacent vertices will not be repeated). To simplify notations, write to denote the binomial defined by the -th column of . Thus . Write , where g_{1}\in\mbox{\mathbb{A}} is a monomial divisible by at least one variable different from . Since there is a directed arc from the vertex to the vertex , then and , where g_{2}\in\mbox{\mathbb{A}} denotes a monomial, possibly constant. Then \mbox{{\rm rad}}(f_{i_{2}},x_{i_{1}})=(x_{i_{2}},x_{i_{1}}). Similarly, \mbox{{\rm rad}}(f_{i_{3}},x_{i_{2}})=(x_{i_{3}},x_{i_{2}}) and so on. Therefore
[TABLE]
∎
We write \mbox{{\rm Hull}}(I) for the intersection of the isolated primary components of (see, e.g., [OP] or [OPV]).
Corollary 2.10**.**
Let be an matrix. Then and I(\mbox{\mathcal{L}}) have height . Moreover, I(\mbox{\mathcal{L}})=\mbox{{\rm Hull}}(I(L)) is the intersection of the isolated primary components of . In particular, I(\mbox{\mathcal{L}}) is a Cohen-Macaulay ideal of dimension and projective dimension .
Proof.
That has height follows from Remark 2.8, Proposition 2.9 and [OPV, Lemma 5.1]. That I(\mbox{\mathcal{L}}) has height and is the Hull of follows from [OPV, Proposition 5.7]. In particular, I(\mbox{\mathcal{L}}) is (homogeneous) unmixed and \mbox{\mathbb{A}}/I(\mbox{\mathcal{L}}) is a (graded) Cohen-Macaulay ring of dimension . By [BH, Corollary 2.2.15], I(\mbox{\mathcal{L}}) is perfect and so the projective dimension of I(\mbox{\mathcal{L}}) is . ∎
3. The Cyc complex associated to a directed graph
3.1. The set of cyclically ordered partitions and an
enumeration
Let us now recall some material from [MaS, Section 2], treating it however in a more general context that the one considered there. Set , take , and let \mbox{{\rm Cyc}}_{n,k} denote the set of cyclically ordered partitions of the set into blocks, which we shall always take to be non-empty. Each element of \mbox{{\rm Cyc}}_{n,k} has the form , where is a partition, and we regard the as formal symbols subject to the identifications
[TABLE]
Each such symbol clearly specifies an equivalence class, and we pick a unique representative of the corresponding equivalence class by assuming, after relabelling of suffices, that .
The cardinality of the set \mbox{{\rm Cyc}}_{n,k} is \lvert\mbox{{\rm Cyc}}_{n,k}\rvert=(k-1)!\cdot S_{n,k}, where is the Stirling number of the second kind. Note that \lvert\mbox{{\rm Cyc}}_{n,1}\rvert=1 and \lvert\mbox{{\rm Cyc}}_{n,n}\rvert=(n-1)!. When the value of is understood from the context, it will be convenient to denote \lvert\mbox{{\rm Cyc}}_{n,k}\rvert by (see Definition 3.3).
We take the opportunity to introduce a convenient enumeration of \mbox{{\rm Cyc}}_{n,k}, which will be used subsequently in the following sections.
Remark 3.1**.**
Given a subset of of cardinality , let denote the incidence vector of , so that \chi_{I}^{\top}=(\chi_{I}(1),\ldots,\chi_{I}(n))\in\{0,1\}^{n}\subset\mbox{\mathbb{Z}}^{n}, with , if , and , otherwise. If and are two different subsets of , we will use the notation and say that “ precedes ” (or alternatively, “ succeeds ”), whenever , or whenever and the difference is ; that is, the rightmost non-zero coefficient of is . Given and , two different cyclically ordered partitions of into blocks, with , we will use the notation and say that “ precedes ” (or alternatively, “ succeeds ”), whenever, for some , and precedes . This is a “size-reverse lexicographic enumeration”, so we will refer to it as an (for a discussion on this subject see, e.g., [LHS]). Any reference made below without qualification to an enumeration will refer to the enumeration employing the . (In this regard, note the word of caution given in Remark 6.18.)
Example 3.2**.**
Set . For the sake of brevity, write for . Understanding that, in the display, elements on the left precede elements on the right and employing the , \mbox{{\rm Cyc}}_{4,k} is enumerated as follows:
[TABLE]
3.2. The Cyc sequence associated to a digraph or to a CB matrix
Let be a digraph of vertices or, equivalently, consider an matrix (recall Assumption 2.2). For disjoint non-empty subsets and of , we define
[TABLE]
If or is the empty set, we put equal to .
Definition 3.3**.**
Let \mbox{\mathcal{C}}_{\mathbb{G}} (also denoted by \mbox{\mathcal{C}}_{L}, or just ) be the sequence of homomorphisms of -modules defined as follows:
[TABLE]
where \mbox{\mathcal{C}}_{k}=\mbox{\mathbb{K}}[x]^{{\rm Cyc}_{n,k+1}} is the free \mbox{\mathbb{K}}[x]-module of rank r_{k}:=\lvert\mbox{{\rm Cyc}}_{n,k+1}\rvert and with basis \mbox{\mathcal{B}}_{k}, the set of elements of \mbox{{\rm Cyc}}_{n,k+1}. The elements of \mbox{\mathcal{B}}_{k} are enumerated employing the . With this enumeration, they will be denoted by \mbox{\mathcal{B}}_{k}=\{e_{k,1},\ldots,e_{k,r_{k}}\}. Their images under \mbox{\partial}_{k} will be denoted by \mbox{\partial}_{k}(\mbox{\mathcal{B}}_{k})=\{f_{k-1,1},\ldots,f_{k-1,r_{k}}\}, with f_{k-1,j}:=\mbox{\partial}_{k}(e_{k,j}). The module \mbox{\mathcal{C}}_{0} is \mbox{\mathbb{K}}[x]^{{\rm Cyc}_{n,1}}=\mbox{\mathbb{K}}[x]. Here the trivial -block will be identified with the unit element of \mbox{\mathbb{K}}[x](=\mbox{\mathbb{A}}). The boundary map \mbox{\partial}_{k}:\mbox{\mathcal{C}}_{k}\to\mbox{\mathcal{C}}_{k-1}, in slightly simplified notation that we use throughout, is given by the formula
[TABLE]
Remark 3.4**.**
Observe that the terms of the first addition in (3.3) are enumerated according to the ; this is not the case for the last term, which depends on the relationship between and . However, provided that, for all , , the last term is distinguished from the others because it is the only one whose coefficient contains the variable . This follows from the fact that .
3.3. The degree zero component of Cyc
Let be a digraph of vertices or, equivalently, consider an matrix . Let be the ideal associated to and let I(\mbox{\mathcal{L}}) be the lattice ideal associated to the lattice spanned by the columns of (see Section 2.1).
Remark 3.5**.**
Let (I,\overline{I})\in\mbox{\mathcal{C}}_{1}, with , and . Let be the incidence vector of . Then
[TABLE]
If follows that I(L)\subseteq\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})\subseteq I(\mbox{\mathcal{L}}) is a chain of inclusions of three ideals of \mbox{\mathbb{K}}[x]=\mbox{\mathbb{A}}. Furthermore, if is an matrix, then , \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) and I(\mbox{\mathcal{L}}) are homogeneous ideals.
Proof.
If , the -th coordinate of is , which is positive. Therefore, in this case, . If , the -th coordinate of is , which is non-positive. Therefore, in this case, . Hence
[TABLE]
which coincides with the definition of . Clearly \chi_{I}+\chi_{\overline{I}}=\mbox{1\mkern-4.6mu{\rm l}}^{\top}. Therefore, and . Thus .
We know that \mbox{\partial}_{1}(I,\overline{I})=x^{I\to\overline{I}}(I\cup\overline{I})-x^{\overline{I}\to I}(I\cup\overline{I}). As we mentioned before, we make the identification with the unit element in \mbox{\mathbb{K}}[x]=\mbox{\mathcal{C}}_{0}. So
[TABLE]
which proves (25). Since L\chi_{I}\in\mbox{\mathcal{L}}, then is in I(\mbox{\mathcal{L}}). This proves \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})\subseteq I(\mbox{\mathcal{L}}).
On the other hand, for , the -th column of can be expressed as . So, by (25), x^{l_{*,i}^{+}}-x^{l_{*,i}^{-}}=\mbox{\partial}_{1}(\{i\},\overline{\{i\}})\in\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}). Recalling that our partitions of have the property , then can be expressed as . Taking into account that, when \beta=-\alpha\in\mbox{\mathbb{Z}}^{n}, then , and using (25) again,
[TABLE]
Therefore, I(L)=(x^{l_{*,1}^{+}}-x^{l_{*,1}^{+}},\ldots,x^{l_{*,n}^{+}}-x^{l_{*,n}^{+}})\subseteq\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}).
Now, suppose that is an matrix. Since L\chi_{I}\in\mbox{\mathcal{L}}, it follows, as in the proof of Remark 2.8, that is homogeneous (in the grading considered in Assumption 2.7). This proves that \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) is homogeneous and \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})\subseteq I(\mbox{\mathcal{L}}). By Remark 2.8, we already know that and I(\mbox{\mathcal{L}}) are homogeneous. ∎
Notation 3.6**.**
For ease of notation, if (C,\overline{C})\in\mbox{\mathcal{C}}_{1} (with , and ) is a partition of , we set , , and f_{C}:=m_{C}-m_{\overline{C}}=\mbox{\partial}_{1}(C,\overline{C}).
Definition 3.7**.**
We define \mbox{\partial}_{0}:\mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x]\to\mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) to be the natural projection onto the quotient ring, so that \mbox{\partial}_{0}\circ\mbox{\partial}_{1}=0.
3.4. The Cyc sequence in four variables
Let us display the sequence of homomorphims of Definition 3.3 for a digraph of four vertices or, equivalently, for a matrix .
Example 3.8**.**
Let . To simplify notations, write instead of , respectively. As in Example 3.2, write
[TABLE]
The resulting complex has the following form.
[TABLE]
The differentials go as follows. Recall that the elements of the basis \mbox{\mathcal{B}}_{k} are enumerated employing the (see Definition 3.3 and Example 3.2; the underlined terms will be the leading ones in an ordering specified subsequently in Assumption 4.9). For \mbox{\partial}_{1}:
[TABLE]
For \mbox{\partial}_{2} we have:
[TABLE]
Finally, for \mbox{\partial}_{3}:
[TABLE]
[TABLE]
3.5. Cyc is a complex for any CB matrix
Let be a matrix or, equivalently, let be a digraph. Let \mbox{\mathcal{C}}_{\mathbb{G}} be the sequence of homomorphisms as in Definition 3.3.
Proposition 3.9**.**
The sequence \mbox{\mathcal{C}}_{\mathbb{G}} is a chain complex of free \mbox{\mathbb{K}}[x]-modules.
Proof.
Fix a basis element (I_{1},I_{2},...,I_{k+1})\in\mbox{\mathcal{B}}_{k} of \mbox{\mathcal{C}}_{k}, where . We need to show that \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},I_{2},...,I_{k+1}))=0. Clearly, \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1})) is expressed as a linear combination of elements of \mbox{\mathcal{C}}_{k-2}. Indeed:
[TABLE]
For , the expression for \mbox{\partial}_{k-1}(I_{1}\cup I_{2},\ldots,I_{k+1}) is equal to:
[TABLE]
For , the expression for \mbox{\partial}_{k-1}(I_{1},I_{2}\cup I_{3},\ldots,I_{k+1}) is:
[TABLE]
For , the expression for \mbox{\partial}_{k-1}(I_{1},\ldots,I_{s}\cup I_{s+1},\ldots,I_{k+1}) is:
[TABLE]
For , the expression for \mbox{\partial}_{k-1}(I_{1},\ldots,I_{k-1}\cup I_{k},I_{k+1}) is equal to:
[TABLE]
For , we have \mbox{\partial}_{k-1}(I_{1},\ldots,I_{k}\cup I_{k+1}) equal to
[TABLE]
As for the final term, \mbox{\partial}_{k-1}(I_{2},\ldots,I_{k},I_{1}\cup I_{k+1}) is equal to:
[TABLE]
The basis elements of \mbox{\mathcal{B}}_{k-2} that appear in these linear combinations, from (31) to (36), solely involve partitions obtained from by coalescing twice two adjacent blocks. Any such basis element appears twice, obtained by doing two different mergings done in two different orders. For instance, the element appears in the first summand in (31) and in (32). In (31), one has first merged with , obtaining and then, subsequently, one has merged with , giving . On the other hand, the first summand in (32) is obtained by first merging with , getting , and afterwards, on merging with , one obtains . Observe that both monomial coefficients are equal, but opposite in sign, so they cancel each with the other.
We list now the complete set of partitions appearing as basis elements in the expression for \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1})). Concretely, and enumerated accordingly to the , the first are:
[TABLE]
These elements appear in (31) and in (32).
Without mentioning where they come from, we have the following, still enumerated with the .
Next we have a set of , all of them beginning with , thus succeeding the above:
[TABLE]
Then we have the set of , beginning with , and so on. In the second last place, we have the set of beginning with , that is, and , and finally we have . Altogether then, these amount in total to partitions.
In the expression for \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1})), there are still further basis elements of \mbox{\mathcal{B}}_{k-2} involved. Concretely, these are the final terms in (31), (32), (33), (34) and (35), and all the terms in (36). It is not possible to enumerate them together with the former ones (employing the ) unless more information is given. Each involves the leftmost block being merged with the rightmost block . These basis elements are also characterized by the fact that the corresponding coefficient term contains the variable (possibly with exponent zero). Enumerated among themselves (in the ), the first are as follows:
[TABLE]
Finally, we have the last term .
Thus a total amount of distinct basis elements of \mbox{\mathcal{B}}_{k-2} appear in \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1})). As mentioned before, each of them appears twice. This is consistent with the total number of elements in the linear expression for \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1})).
To finish the proof, one can check that the two terms involving each basis element have the same monomial coefficient, but with opposite sign (alternatively, one can use (41) below). This will prove that \mbox{\partial}_{k-1}(\mbox{\partial}_{k}(I_{1},\ldots,I_{k+1}))=0.
- •
Terms , with . The two monomial coefficients involving this basis element are:
[TABLE]
- •
Terms , with , and . The two monomial coefficients involving this basis element are:
[TABLE]
- •
Terms , with . Here, the two monomial coefficients are:
[TABLE]
- •
Terms . The two monomial coefficients are:
[TABLE]
- •
Finally, the two monomial coefficients of are
[TABLE]
It is easy to check that they are equal in pairs, but opposite in sign. This finishes the proof. ∎
Remark 3.10**.**
If is strongly complete, or equivalently is a matrix, then \mbox{\partial}_{k}(\mbox{\mathcal{C}}_{k})\subseteq\mbox{\mathfrak{m}}\mbox{\mathcal{C}}_{k-1}, since , for all , ensures the containment x^{I_{t}\to I_{t+1}}\in(x_{1},\ldots,x_{n})=\mbox{\mathfrak{m}}.
4. Gröbner bases for lattice ideals associated to
strongly connected digraphs
4.1. Block echelon form of an ICB matrix
We start by defining what we will consider to be a block echelon form of an matrix. In order to simplify the exposition, the expression “-rows” will stand for the set of rows whose row subindex is in , and analogously as regards the expression “-columns”. As in (8), given , let be the submatrix of defined by the -rows and the -columns.
Definition 4.1**.**
Let be an matrix and let be a positive integer, . Given q_{1},\ldots,q_{\delta}\in\mbox{\mathbb{N}}_{+}, with , let be the partition of into blocks of cardinalities , defined by , . Set . Then is said to be in -block echelon form if , for every and such that and , and each column of is nonzero, for every .
Example 4.2**.**
Let be the matrix below which is in -block echelon form, where , and , and . Then . Note that the blocks on the diagonal have diagonal positive entries. The blocks below the diagonal blocks have nonzero columns. All the remaining blocks below the aforementioned are zero.
[TABLE]
Remark 4.3**.**
Any matrix can be reduced by permutations to a -block echelon form.
Proof.
(We remark that it may help to consider the shape of the matrix in the previous example.)
Indeed, set . The notation -columns, with , will denote the set of columns with column subindex in , having a nonzero entry in the last row. Since is irreducible, then, necessarily, in the last row of there are other nonzero entries apart from (this follows too from the fact that since is, in particular, a matrix, ). Therefore . Move these columns to the outer right of the first columns; we use the notation -columns to denote the corresponding set of columns. Then perform the corresponding permutation of rows, so preserving the structure of a matrix, while in the last row, which remains untouched by the row permutations just carried out, we maintain the property that all its nonzero entries are in the set of -columns.
Since is irreducible, then there must be at least one column to the left of the set of -columns having a nonzero entry in the set of -rows, these being strictly below the diagonal. For, if not, setting and , one would have a partition of into two non-empty disjoint sets and such that . Thus, would be reducible, a contradiction (recall the definition of irreducible in Section 2.3). So, denote by the set of column indices to the left of the set of -columns which have a nonzero entry in the set of -rows, these being strictly below the diagonal, where . Now proceed recursively. ∎
Remark 4.4**.**
Let be an matrix in -block echelon form. Let , . Then the rightmost nonzero coefficient of is negative.
Proof.
Keeping the notations in Definition 4.1, set and let be the maximum integer with . Clearly and . Since all the columns to the left of have zero -rows, it is enough to see that and different from zero. This follows from the fact that and that each column of is nonzero. ∎
To finish this section, let us prove that the -block-echelon form reflects the natural partition of the set of vertices into subsets of elements whose distance from the last vertex is constant. This brings out another aspect of the relationship between matrices and -block echelon form, this time arising from the natural metric on \mbox{\mathbb{G}}_{L}.
Lemma 4.5**.**
Let be an matrix and let \mbox{\mathbb{G}}_{L}=(\mbox{\mathbb{V}},\mbox{\mathbb{E}}) its associated digraph. Suppose that is in -block echelon form and let be the associated partition of into blocks, with . For , set W_{i}:=\{v_{k}\in\mbox{\mathbb{V}}\mid k\in I_{i}\}, so that . Then is a partition of \mbox{\mathbb{V}}\setminus\{v_{n}\} into non-empty subsets of . Moreover, for each , the set coincides with the set V_{i}:=\{v\in\mbox{\mathbb{V}}\mid\mbox{{\rm d}^{\vee}}(v_{n},v)=\delta+1-i\}.
Proof.
Clearly, since is a partition of into non-empty blocks, then is a partition of \mbox{\mathbb{V}}\setminus\{v_{n}\} into non-empty subsets of . Let us prove that by decreasing induction on . Clearly .
Fix , with . Suppose that and let us prove .
Take , so that . Since has non-zero columns, there exists , with , so there is an arc in . Similarly, since has non-zero columns, there exists , with , and an arc in . Recursively, there are vertices and a directed path . In particular, \mbox{{\rm d}^{\vee}}(v_{n},v_{k_{i}})\leq\delta+1-i and . On the other hand, since is a partition of and , then . Thus and it follows that .
Now, take . Then there exist , a directed path of minimum unweighted length . In particular, \mbox{{\rm d}^{\vee}}(v_{n},v_{k_{i+1}})\leq\delta-i and, in fact, this is an equality, otherwise \mbox{{\rm d}^{\vee}}(v_{n},v_{k})<\delta+1-i. Hence and . The existence of the arc ensures . Since is in -block echelon form, the index must sit inside . If , then , a contradiction with the fact that is clearly disjoint with . Thus and and we derive the opposite inclusion . ∎
4.2. A convenient enumeration of the vertices
Taking into acccount the former lemma, let us define a special enumeration on the set of vertices and arcs of a strongly connected digraph, and hence on the set of indeterminates (see [CRS, p. 7], [MoS, Section 2.2.3], for the case where is undirected, or, equivalently, is symmetric).
Definition 4.6**.**
Let \mbox{\mathbb{G}}=(\mbox{\mathbb{V}},\mbox{\mathbb{E}}) be a strongly connected digraph. Fix a distinguished vertex in . Since is finite, \delta:=\max\{\mbox{{\rm d}^{\vee}}(\omega,v)\mid v\in\mbox{\mathbb{V}}\} is finite and . Consider the partition of \mbox{\mathbb{V}}\setminus\{\omega\} into non-empty subsets defined by V_{i}=\{v\in\mbox{\mathbb{V}}\mid\mbox{{\rm d}^{\vee}}(\omega,v)=\delta+1-i\}, for . Set . Let be the cardinality of each . Take the partition of into blocks of cardinalities , defined by , . Set . The -enumeration of the vertices and arcs of is determined by assigning the set of indices to the set of vertices . That is, V_{i}=\{v_{k}\in\mbox{\mathbb{V}}\mid k\in I_{i}\}, for . The index is assigned to the vertex , so that becomes . In particular, if , then \mbox{{\rm d}^{\vee}}(v_{n},v_{i})\geq\mbox{{\rm d}^{\vee}}(v_{n},v_{j}). On the other hand, if \mbox{{\rm d}^{\vee}}(v_{n},v_{i})>\mbox{{\rm d}^{\vee}}(v_{n},v_{j}), then . Clearly, whenever \mbox{{\rm d}^{\vee}}(v_{n},v_{i})=\mbox{{\rm d}^{\vee}}(v_{n},v_{j}), the three possibilities , or may happen.
In Lemma 4.5 we have seen that if is an matrix in -block echelon form, then the corresponding digraph \mbox{\mathbb{G}}_{L} has an -enumeration (where is taken to be the last vertex). Now let us prove that a digraph with an -enumeration has the corresponding matrix in -block echelon form.
Lemma 4.7**.**
Let \mbox{\mathbb{G}}=(\mbox{\mathbb{V}},\mbox{\mathbb{E}}) be a strongly connected digraph with an -enumera-tion. Let be the matrix associated to . Then is in -block echelon form.
Proof.
With the notation in Definition 4.6,
[TABLE]
for , and and . We want to prove that , for every and such that and , and each column of is nonzero, for every .
So, fix with and . Then, we need to prove that
[TABLE]
Note that, for , then and , for all , since the vertices are such that \mbox{{\rm d}^{\vee}}(\omega,v_{k})=1.
Suppose that for some . Let be a directed path of minimum unweighted length , where (because ). Since , there is an arc in , and so a directed path from to of unweighted length at most , a contradiction, since \mbox{{\rm d}^{\vee}}(v_{n},v_{k})=\delta+1-j (recall that ). This proves the first part of (38).
Now, take a directed path of minimum unweighted length . In particular, \mbox{{\rm d}^{\vee}}(\omega,v_{i_{\delta-j}})=\delta-j. Thus and , since is an arc of . This proves the second part of (38). ∎
Assumption 4.8**.**
Let be an matrix or, equivalently, let \mbox{\mathbb{G}}_{L} be a strongly connected digraph. From now on, we will assume that is always in -block echelon form or, equivalently, that has a -enumeration (see Lemmas 4.5 and 4.7).
Note that, if is a matrix or, equivalently, is strongly complete, this assumption is vacuous.
4.3. The weighted reverse lexicographic order on the polynomial
ring
Let be an matrix or, equivalently, let \mbox{\mathbb{G}}_{L} be a strongly connected digraph. Recall that, by Assumption 2.7, and .
Assumption 4.9**.**
From now on, we will always suppose that \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x]=\mbox{\mathbb{K}}[x_{1},\ldots,x_{n}] is endowed with the weighted reverse lexicographic order (, for short). Concretely, given \alpha,\beta\in\mbox{\mathbb{N}}^{n}, we set precisely when either , or and there exists such that and . This last condition will be written . See, e.g, [GP, p. 13], where this ordering is denoted by .
We insert a note of caution: it need no longer be the case that, under this , , since, for example, we have whenever .
Given a non-zero polynomial f\in\mbox{\mathbb{A}}, we let \mbox{{\bf L}{\rm c}}(f), \mbox{{\bf L}{\rm m}}(f) and \mbox{{\bf L}{\rm t}}(f) denote the leading coefficient, the leading monomial and the leading term, respectively, of with respect to the (see, e.g., [CLO, § 2.2, Definition 7]).
Recall that f_{C}:=m_{C}-m_{\overline{C}}=\mbox{\partial}_{1}(C,\overline{C}), where , (see Notation 3.6).
Lemma 4.10**.**
Let (C,\overline{C})\in\mbox{\mathcal{C}}_{1} and set f_{C}:=m_{C}-m_{\overline{C}}=\mbox{\partial}_{1}(C,\overline{C}). Then is homogeneous and \mbox{{\bf L}{\rm m}}(f_{C})=m_{C}. In particular, \mbox{{\bf L}{\rm t}}(f_{C})=\mbox{{\bf L}{\rm m}}(f_{C}).
Proof.
That is homogeneous is shown in Remark 3.5. So \mbox{{\rm grade}}(m_{C})=\mbox{{\rm grade}}(x^{C\to\overline{C}}) is equal to \mbox{{\rm grade}}(m_{\overline{C}})=\mbox{{\rm grade}}(x^{\overline{C}\to C}). Also by Remark 3.5, we know that, and , where is the incidence vector of (see Remark 3.1). Then . So we must see that the rightmost nonzero coefficient of is negative. But this follows from Remark 4.4. ∎
If the variables are not enumerated according to Assumption 4.8, then Lemma 4.10 may fail.
Example 4.11**.**
Consider the following matrices and :
[TABLE]
[TABLE]
Observe that arises from by the transposition of rows and columns and of . The corresponding digraphs \mbox{\mathbb{G}}_{L} and \mbox{\mathbb{G}}_{L^{\prime}} are the same, but with a re-enumeration of vertices and arcs. Since \mbox{\mathbb{G}}_{L} and \mbox{\mathbb{G}}_{L^{\prime}} are strongly connected, and are matrices. Note that is in -block echelon form, whereas is not.
A simple check shows that
[TABLE]
Imposing the usual on our polynomial ring \mbox{\mathbb{K}}[x,y,z,t], with being replaced by , respectively, then , , and , when working with . Considering and , we have: , , . Note that because , which is consistent with . Thus \mbox{{\bf L}{\rm m}}(f_{C})=x^{2}=m_{\overline{C}}.
4.4. A Gröbner basis for the image of the first differential
Let be an matrix or, equivalently, let \mbox{\mathbb{G}}_{L} be a strongly connected digraph. Recall that \mbox{\mathcal{B}}_{k} is the set of elements of \mbox{{\rm Cyc}}_{n,k+1}, enumerated employing the (see Definition 3.3), forming a basis of the free \mbox{\mathbb{K}}[x]-module \mbox{\mathcal{C}}_{k}=\mbox{\mathbb{K}}[x]^{{\rm Cyc}_{n,k+1}}. Concretely, in degree zero:
[TABLE]
The purpose of this subsection is to prove that \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a Gröbner basis of \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) (see also Notation 3.6). We begin with some useful notations and remarks.
Notation 4.12**.**
Given three disjoint subsets , note that:
[TABLE]
We set:
[TABLE]
In particular,
[TABLE]
Notation 4.13**.**
Let and be two non-empty subsets of . Write , , , and . In particular, and . The following picture may help in reading everything that will follow.
[TABLE]
Lemma 4.14**.**
The restriction of the map \mbox{\partial}_{1}:\mbox{\mathcal{C}}_{1}\to\mbox{\mathcal{C}}_{0} to the set \mbox{\mathcal{B}}_{1} is injective. Concretely, given , two non-empty subsets of , then if and only if .
Proof.
Suppose that . By Lemma 4.10, m_{C}=\mbox{{\bf L}{\rm t}}(f_{C})=\mbox{{\bf L}{\rm t}}(f_{D})=m_{D}. Thus and hence . Suppose that . Since the initial hypotheses are symmetrical in and , we may assume without loss of generality that , that is, that , with Notation 4.13 in force. Using (41), one has:
[TABLE]
and
[TABLE]
It follows from the equality that , i.e., that , for all and . We deduce that from the equality , i.e., that , for all and . Thus , for all and . Therefore there is no directed path joining any vertex in to the vertex , which is a contradiction to the hypothesis that \mbox{\mathbb{G}}_{L} is a strongly connected digraph. ∎
Let us calculate the -polynomial of f_{C}=\mbox{\partial}_{1}(C,\overline{C}) and f_{D}=\mbox{\partial}_{1}(D,\overline{D}). Observe that if , then and .
Lemma 4.15**.**
Let and be two non-empty distinct subsets of . Consider the monomials and , defined as follows:
[TABLE]
Suppose that and that . Then are non-empty subsets of , so and are in \mbox{\mathcal{C}}_{1}. Then the -polynomial of and is
[TABLE]
with f_{F}=\mbox{\partial}_{1}(F,\overline{F}) and f_{G}=\mbox{\partial}_{1}(G,\overline{G}). Moreover,
[TABLE]
Suppose, without loss of generality, that . Then and is a non-empty subset of , so (G,\overline{G})\in\mbox{\mathcal{C}}_{1}. Then the -polynomial of and is
[TABLE]
with f_{G}=\mbox{\partial}_{1}(G,\overline{G}). Moreover, \mbox{{\bf L}{\rm m}}(S(f_{C},f_{D}))\geq\mbox{{\bf L}{\rm m}}(l_{D,C}f_{G}).
Proof.
By definition (see, e.g., [CLO, § 2.6, Definition 4]), , where
[TABLE]
Suppose that and , so that . By Lemma 4.10, \mbox{{\bf L}{\rm t}}(f_{C})=\mbox{{\bf L}{\rm m}}(f_{C})=m_{C}=x^{C\to\overline{C}} and \mbox{{\bf L}{\rm t}}(f_{D})=\mbox{{\bf L}{\rm m}}(f_{D})=m_{D}=x^{D\to\overline{D}}. Using (41) and (43), and the fact that \mbox{{\rm LCM}}(fg,fh)=f\cdot\mbox{{\rm LCM}}(g,h) for elements f,g,h\in\mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x], one has:
[TABLE]
In a symmetric manner,
[TABLE]
Set and . Therefore and . Analogously, and . Then
[TABLE]
Set
[TABLE]
so that . An easy computation using (41) shows that
[TABLE]
However, using (41) and then (43), we get
[TABLE]
Therefore, . On the other hand,
[TABLE]
Set
[TABLE]
so that . An easy computation using (41) shows that
[TABLE]
The situation is symmetrical in and . Swapping and in an expression forces a swap in and , with , and invariant. Thus follows by symmetry from the equality (50). Therefore, and . An easy computation using (41) shows that and, by symmetry, . Therefore, .
Observe that one can consider (F,\overline{F}),(G,\overline{G})\in\mbox{\mathcal{C}}_{1} due to the hypothesis . Using Lemma 4.10, it follows that \mbox{{\bf L}{\rm m}}(f_{F})=m_{F} and \mbox{{\bf L}{\rm m}}(f_{G})=m_{G}. Since and are monomials, then \mbox{{\bf L}{\rm m}}(l_{C,D}f_{F})=l_{C,D}m_{F}=m_{5} and \mbox{{\bf L}{\rm m}}(l_{D,C}f_{G})=l_{D,C}m_{G}=m_{7} (see, e.g., [CLO, §2.2, Exercise 11]). On the other hand, . Thus, one of the two monomials, either , or else , is the leading monomial of . If , then \mbox{{\bf L}{\rm m}}(l_{C,D}f_{F})=m_{5}=m_{4}=\mbox{{\bf L}{\rm m}}(S(f_{C},f_{D})) and \mbox{{\bf L}{\rm m}}(l_{D,C}f_{G})=m_{7}=m_{2}<m_{4}=\mbox{{\bf L}{\rm m}}(S(f_{C},f_{D})). On the contrary, if , then \mbox{{\bf L}{\rm m}}(l_{C,D}f_{F})=m_{5}=m_{4}<m_{2}=\mbox{{\bf L}{\rm m}}(S(f_{C},f_{D})) and \mbox{{\bf L}{\rm m}}(l_{D,C}f_{G})=m_{7}=m_{2}=\mbox{{\bf L}{\rm m}}(S(f_{C},f_{D})).
Finally, assume, without loss of generality, that . Then and . Hence and ; and ; and . As before, and . On the other hand, since , then and , which is readily seen to be equal to . Moreover, trivially then, one has \mbox{{\bf L}{\rm m}}(S(f_{C},f_{D}))\geq\mbox{{\bf L}{\rm m}}(l_{D,C}f_{G}). ∎
Proposition 4.16**.**
The set \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1})=\{f_{C}\mid C\subseteq[n-1],C\neq\varnothing\} is a Gröbner basis for the ideal \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}).
Proof.
This follows from Lemma 4.15 and the criterion of Buchberger (see, e.g., [CLO, § 2.6, Theorem 6 and § 2.9, Theorem 3]). ∎
4.5. The Cyc complex is exact in degree zero
Let be an matrix or, equivalently, let be a strongly connected digraph. Recall that \mbox{\partial}_{0}:\mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x]\to\mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) is defined to be the natural projection onto the quotient ring, so that \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})\subseteq I(\mbox{\mathcal{L}}) (Definition 3.7).
Proposition 4.17**.**
We have \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}):x_{n}=\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) and \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})=I(\mbox{\mathcal{L}})=\ker(\mbox{\partial}_{0}).
Proof.
According to Definition 3.3, let
[TABLE]
enumerated employing the . By Proposition 4.16, \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a Gröbner basis for the ideal \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}). Now, take f\in\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}):x_{n}. Using the Division Algorithm, can be written as
[TABLE]
where either , or else and no term in is divisible by any one of \mbox{{\bf L}{\rm t}}(f_{0,i}), for (see, e.g., [CLO, § 2.3, Theorem 3]). Suppose that . On multiplying by in (51), one deduces that x_{n}h\in\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) and so \mbox{{\bf L}{\rm t}}(x_{n}h)\in\mbox{{\bf L}{\rm t}}(\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})). Since is a Gröbner basis of \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}), \mbox{{\bf L}{\rm t}}(\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}))=\langle\mbox{{\bf L}{\rm t}}(f_{0,1}),\ldots,\mbox{{\bf L}{\rm t}}(f_{0,r_{1}})\rangle. By, e.g., [CLO, § 2.4, Lemma 2], \mbox{{\bf L}{\rm t}}(x_{n}h)=x_{n}\mbox{{\bf L}{\rm t}}(h) is a multiple of some \mbox{{\bf L}{\rm t}}(f_{0,_{i}}). Since does not occur in any \mbox{{\bf L}{\rm t}}(f_{0,i}), it follows that \mbox{{\bf L}{\rm t}}(h) is a multiple of some \mbox{{\bf L}{\rm t}}(f_{0,i}), a contradiction. Therefore \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}):x_{n}=\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}). In particular, \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}):x_{n}^{m}=\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}), for all .
By Remark 3.5, I(L)\subseteq\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})\subseteq I(\mbox{\mathcal{L}}). By [OPV, Proposition 5.7], I(\mbox{\mathcal{L}})=I(L):x_{n}^{m}, for some . Therefore
[TABLE]
so I(\mbox{\mathcal{L}})=\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) as desired. ∎
Putting together Propositions 4.16 and 4.17, we get the following result.
Corollary 4.18**.**
The set \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1})\subset\mbox{\mathcal{C}}_{0} is a Gröbner basis of \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})=I(\mbox{\mathcal{L}})=\ker(\mbox{\partial}_{0}).
4.6. Minimality in the strongly complete case
Let be an matrix or, equivalently, let be a strongly connected digraph.
Proposition 4.19**.**
Suppose that is strongly complete. Then \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a minimal Gröbner basis and a minimal homogeneous system of generators of I(\mbox{\mathcal{L}}).
Proof.
Recall that \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1})=\{f_{0,1},\ldots,f_{0,r_{1}}\}=\{f_{C}\mid C\subseteq[n-1],C\neq\varnothing\}. By Proposition 4.18, \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a Gröbner basis of I(\mbox{\mathcal{L}}). Let us see that it is a minimal such basis (see, e.g., [CLO, § 2.7. Definition 4]). Fix , , . Clearly, the leading coefficient of is . Moreover, the leading term of is \mbox{{\bf L}{\rm t}}(f_{C})=m_{C}=x^{C\to\overline{C}} (see Remark 3.6 and Lemma 4.10). Suppose that \mbox{{\bf L}{\rm t}}(f_{C}) lies in the ideal \langle\mbox{{\bf L}{\rm t}}(f_{D})\mid D\subseteq[n-1],D\neq C,D\neq\varnothing\rangle. Since the latter is a monomial ideal, it follows that \mbox{{\bf L}{\rm t}}(f_{D}) divides \mbox{{\bf L}{\rm t}}(f_{C}), for some , , . Now, use that is strongly complete, or equivalently, is a matrix. Therefore, , for all . Since
[TABLE]
and , it follows that , and so is a proper subset of . However, if , then the exponent of in is , whereas the exponent of in is , which is strictly smaller, because , a contradiction.
By Corollary 4.18, \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a homogeneous system of generators for \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})=I(\mbox{\mathcal{L}}). Let us see that it is a minimal one. Suppose that
[TABLE]
for some polynomials u_{i}\in\mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x]. In other words, (u_{1},\ldots,u_{r_{1}})\in\mbox{\mathbb{A}}^{r_{1}} is in the kernel of \varphi:\mbox{\mathbb{A}}^{r_{1}}\to\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}), the free -linear presentation of \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}), sending each element of the canonical basis to . Write each in the form , v_{i},w_{i}\in\mbox{\mathbb{K}}[x], where contains no term involving . Set in the equality (52) above. Then we get
[TABLE]
so is a syzygy on the leading terms \mbox{{\bf L}{\rm t}}(f_{0,1}),\ldots,\mbox{{\bf L}{\rm t}}(f_{0,r_{1}}). By [CLO, § 2.9. Proposition 8, see also Definition 5], is of the form , where
[TABLE]
We have and , for some , and . Then
[TABLE]
with the notations as in the equality (47). Writing , , , and and using the equalities (4.4) and (49) in the proof of Lemma 4.15, then
[TABLE]
Suppose that and . Then and are non-empty. Since , for all , we deduce that and , and so and , are in the irrelevant maximal ideal . Suppose, without loss of generality, that . Then and , where , and are non-empty (see the last paragraph in the proof of Lemma 4.15). Again, we deduce that and are in . Hence each also lies in and (u_{1},\ldots,u_{r_{1}})\in\mbox{\mathfrak{m}}\mbox{\mathbb{A}}^{r_{1}}. On tensoring the short exact sequence
[TABLE]
by \mbox{\mathbb{A}}/\mbox{\mathfrak{m}}=k, we deduce that the induced map \overline{\varphi}:\mbox{\mathbb{A}}^{r_{1}}/\mbox{\mathfrak{m}}\mbox{\mathbb{A}}^{r_{1}}\to\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})/\mbox{\mathfrak{m}}\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) is an isomorphism of -vector spaces. Applying the graded Nakayama Lemma, we deduce that \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a minimal homogeneous system of generators for \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}) (see, e.g., [BH, Exercise 1.5.24]). ∎
Remark 4.20**.**
The minimality for the strongly complete case will also follow from Corollary 6.16, since \ker(\mbox{\partial}_{1})=\mbox{\partial}_{2}(\mbox{\mathcal{C}}_{2})\subset\mbox{\mathfrak{m}}\mbox{\mathcal{C}}_{1} and so \mbox{\mathcal{C}}_{1}/\mbox{\mathfrak{m}}\mbox{\mathcal{C}}_{1}\cong\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})/\mbox{\mathfrak{m}}\mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1}).
4.7. Non-minimality in the non-strongly complete case
Let be an matrix or, equivalently, let be a strongly connected digraph. It is easy to give an example where, if is not strongly complete, then \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is not a minimal system of generators of I(\mbox{\mathcal{L}}).
Example 4.21**.**
Consider the following matrix , which is not a matrix:
[TABLE]
Then \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1})=\{x-t,y-t,xz-yt,x-z,z-t,y-z,x-y\}, whereas , which is clearly minimally generated by . Thus is a complete intersection ideal. In particular, it is unmixed and I(L)=I(\mbox{\mathcal{L}}). Therefore \mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) cannot be a minimal homogeneous system of generators of I(\mbox{\mathcal{L}}).
5. Computing syzygies of a Gröbner basis
5.1. Preliminary notations
Let us recall some preliminary notations. We follow [EMSS, Section 2], [BS] and [E, § 15].
Definitions 5.1**.**
Let F=\mbox{\mathbb{A}}^{r} be the free \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x]-module of rank , and let be the canonical basis of .
Monomials. A monomial in is the product of a monomial in \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x] with a basis element . A term in is the product of a monomial in with a scalar in .
Divisibility. A monomial divides a monomial if and divides ; in this case, is defined as . Similarly, a monomial divides if divides and in this case is defined as . The least common multiple of two monomials and of is \mbox{{\rm LCM}}(m_{1}e_{i},m_{2}e_{j})=\mbox{{\rm LCM}}(m_{1},m_{2})e_{i}, if , or [math] otherwise.
Orderings. A (global) monomial ordering on is a total ordering on the set of monomials of such that if and are monomials in and is a monomial in , then . Moreover, for all and all monomials , and we require in addition that , for all .
Leading terms. Let be a monomial ordering on and , , where . Then the leading coefficient, the leading monomial and the leading term of are \mbox{{\bf L}{\rm c}}(f)=\lambda, \mbox{{\bf L}{\rm m}}(f)=me_{i} and \mbox{{\bf L}{\rm t}}(f)=\lambda me_{i}, respectively. For any subset , the leading module of is defined as \mbox{{\bf L}}(S):=\langle\mbox{{\bf L}{\rm m}}(f)\mid f\in S\setminus\{0\}\rangle.
Definitions 5.2**.**
Let F_{0}=\mbox{\mathbb{A}}^{s} be the free -module of rank and let be a finite subset of . Take F_{1}=\mbox{\mathbb{A}}^{r}, the free -module of rank , the cardinality of , and let be defined by .
Syzygies. The (first) syzygy module of is \mbox{{\rm Syz}}(G_{0}):=\ker(\varphi_{1}). An element of is called a syzygy of .
Induced orderings. Given a monomial ordering on , the induced ordering on (w.r.t. and ) is the monomial ordering defined by m_{1}e_{i}>m_{2}e_{j}\Leftrightarrow\mbox{{\bf L}{\rm m}}(m_{1}f_{i})>\mbox{{\bf L}{\rm m}}(m_{2}f_{j}), or \mbox{{\bf L}{\rm m}}(m_{1}f_{i})=\mbox{{\bf L}{\rm m}}(m_{2}f_{j}) and .
The -vectors. For , the -vector of and is defined as
[TABLE]
In particular, \mbox{{\bf L}{\rm t}}(m^{1}_{j,i}f_{i})=m^{1}_{j,i}\mbox{{\bf L}{\rm t}}(f_{i})=\mbox{{\rm LCM}}(\mbox{{\bf L}{\rm m}}(f_{j}),\mbox{{\bf L}{\rm m}}(f_{i})), which, by symmetry, will be equal to \mbox{{\bf L}{\rm t}}(m^{1}_{i,j}f_{j}). Therefore \mbox{{\bf L}{\rm m}}(m^{1}_{j,i}f_{i})=\mbox{{\bf L}{\rm m}}(m^{1}_{i,j}f_{j})>\mbox{{\bf L}{\rm m}}(S(f_{i},f_{j})).
Standard expression. Let . An equality , with g_{i}\in\mbox{\mathbb{A}} and , is a standard expression for with remainder (and w.r.t. and ) if the following two conditions are satisfied:
\mbox{{\bf L}{\rm m}}(g)\geq\mbox{{\bf L}{\rm m}}(g_{i}f_{i}), for all , whenever both and are nonzero.
If is nonzero, then \mbox{{\bf L}{\rm t}}(h) is not divisible by any \mbox{{\bf L}{\rm t}}(f_{i}).
Discussion 5.3**.**
Let F_{0}=\mbox{\mathbb{A}}^{s} be the free -module of rank and let be a finite subset of . Take F_{1}=\mbox{\mathbb{A}}^{r}, the free -module of rank , the cardinality of , and let be defined by . Suppose that is a Gröbner basis w.r.t. some ordering . By the Buchberger Criterion (see, e.g., [E, Theorem 15.8]), there are standard expressions with remainder zero:
[TABLE]
In particular, \mbox{{\bf L}{\rm m}}(m^{1}_{j,i}f_{i})=\mbox{{\bf L}{\rm m}}(m^{1}_{i,j}f_{j})>\mbox{{\bf L}{\rm m}}(S(f_{i},f_{j}))\geq\mbox{{\bf L}{\rm m}}(g_{s}^{(i,j)}f_{s}), for all , whenever both and the particular are nonzero.
Each such expression (54) defines a syzygy of in , associated to the -vector of and :
[TABLE]
This element will be called a -syzygy associated to the -vector of and . Note that, unless the standard expression (54) is a “determinate division with reminder”, the elements are not necessarily uniquely determined (see [BS, Theorem 1.3], [DS, Theorem 2.2.12]).
However, if , since \mbox{{\bf L}{\rm m}}(m^{1}_{j,i}f_{i})=\mbox{{\bf L}{\rm m}}(m^{1}_{i,j}f_{j}), then . Moreover, since \mbox{{\bf L}{\rm m}}(m^{1}_{i,j}f_{j})>\mbox{{\bf L}{\rm m}}(S(f_{i},f_{j}))\geq\mbox{{\bf L}{\rm m}}(g_{s}^{(i,j)}f_{s}), then m^{1}_{i,j}e_{j}>\mbox{{\bf L}{\rm m}}(g_{s}^{(i,j)})e_{s}, for all , whenever the particular is nonzero.
Thus, if is a -syzygy associated to the -vector of two elements of , obtained as described above, setting (respectively, ) to be the unique basis element involved in \mbox{{\bf L}{\rm t}}(\sigma)/\mbox{{\bf L}{\rm c}}(\sigma) (respectively, \mbox{{\bf L}{\rm t}}(\sigma-\mbox{{\bf L}{\rm t}}(\sigma))/\mbox{{\bf L}{\rm c}}(\sigma-\mbox{{\bf L}{\rm t}}(\sigma))), one deduces that is a -syzygy associated to the -vector of and , with . In particular, \mbox{{\bf L}{\rm t}}(\sigma)=m^{1}_{j,i}e_{i} (see [BS, p.6], [DS, p.68] or [EMSS, Remark 3.6]).
A -syzygy associated to the -vector of and will be denoted by . However, this notation can be somewhat misleading, in the following sense. Suppose that are four elements of , with . The equality does not ensure is equal to , because of the non-uniqueness in the choice of the elements (unless “determinate division with remainder” is used, as mentioned above). Nevertheless, the equality does ensure that and that , as seen above.
For ease of reference, we highlight the property:
[TABLE]
5.2. The Schreyer algorithm to compute
the syzgygies of a Gröbner basis
We recall the Schreyer algorithm to compute the syzygies of a Gröbner basis (see [EMSS, Theorem 3.2 and Proposition 3.5], [DS, Corollary 2.3.19], [BS, Corollary 1.11], [E, Theorem 15.10, Corollary 15.11]). We will refer to [EMSS] particularly for easy quoting. First we introduce some more notations.
Notation 5.4**.**
Let F_{0}=\mbox{\mathbb{A}}^{s} be the free -module of rank and let be a finite subset of . Recall that m^{1}_{j,i}=\mbox{{\rm LCM}}(\mbox{{\bf L}{\rm m}}(f_{j}),\mbox{{\bf L}{\rm m}}(f_{i}))/\mbox{{\bf L}{\rm t}}(f_{i}). Let be the set of ideals of \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x] defined as follows: , and for each ,
[TABLE]
Note that, for each ,
[TABLE]
Therefore, . Observe that, in fact, the are monomial ideals.
Theorem 5.5**.**
([EMSS, Proposition 3.5])* Let F_{0}=\mbox{\mathbb{A}}^{s} be the free -module of rank and let be a finite subset of . Suppose that is a Gröbner basis of w.r.t. some monomial ordering on . Let F_{1}=\mbox{\mathbb{A}}^{r} be the free -module of rank , the cardinality of , and with canonical basis . Let be defined by . For each , and for each minimal generator of , choose exactly one index , with , such that divides , and compute a standard expression for with remainder zero:*
[TABLE]
Let be the corresponding -syzygy associated to the -vector of and . Then
[TABLE]
is a Gröbner basis of w.r.t. the monomial ordering on induced by and . In particular, .
Remark 5.6**.**
Fix . As remarked above, . By definition, is a minimal generator of . Hence if we choose an index with such that the generator of divides , then necessarily equals up to a nonzero scalar. In the situation where the are monomials up to , which will be the case in our applications of Theorem 5.5, then .
Remark 5.7**.**
Suppose that we have a -syzygy such that is a non-minimal generator of and let be as in Theorem 5.5. Then
[TABLE]
is still a Gröbner basis of . This follows from the definition of Gröbner basis. Indeed, since is a Gröbner basis, then \mbox{{\bf L}}(G_{1})=\mbox{{\bf L}}(\ker(\varphi_{1})). Since , then \mbox{{\bf L}{\rm t}}(\tau_{1}(e_{i},e_{j}))\in\mbox{{\bf L}}(\ker(\varphi_{1})) and \mbox{{\bf L}}(\tilde{G}_{1}) is still equal to \mbox{{\bf L}}(\ker(\varphi_{1})).
Remark 5.8**.**
We will consider in the following enumeration: let and be two elements of , with , such that . That is, (cf. Discussion 5.3). We will use the notation and say that “ precedes ”, or alternatively “ succeeds ”, whenever , or whenever and .
6. Exactness of the Cyc complex
6.1. Statement of the main result
The main result of the paper is the following.
Theorem 6.1**.**
Let be a strongly connected digraph of vertices or, equivalently, let be an matrix. Let \mbox{\mathcal{C}}_{\mathbb{G}} be the associated complex. Then
[TABLE]
is a free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}).
Remark 6.2**.**
Recall that we assume that is a finite, weighted, directed graph, without loops, sources or sinks (cf. Assumption 2.2). In particular, its Laplacian matrix is a matrix and, since is strongly connected, then is an matrix (see Subsection 2.5). By Proposition 3.9, \mbox{\mathcal{C}}_{\mathbb{G}} is a chain complex of free -modules. Moreover, we suppose in \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x] the -grading given by (cf. Notation 2.4 and Assumption 2.7). Furthermore, we assume that has a -enumeration or, equivalently, that is in -block echelon form (see 4.8). Finally, we suppose that \mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x] is endowed with the (Assumption 4.9).
Recall also that \mbox{\partial}_{0}:\mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x]\to\mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) is defined to be the natural projection onto the quotient ring (see Definition 3.7). Observe that we may interpret \mbox{\mathcal{C}}_{n}=0 and \mbox{\partial}_{n}=0.
Let us define a monomial ordering on each module \mbox{\mathcal{C}}_{k} of the complex.
Remark 6.3**.**
Recall that the elements of the enumerated free basis \mbox{\mathcal{B}}_{k} of the free module \mbox{\mathcal{C}}_{k} are denoted by , where r_{k}=\mbox{{\rm rank}}(\mbox{\mathcal{C}}_{k})=\lvert\mbox{{\rm Cyc}}_{n,k+1}\rvert. Their images under \mbox{\partial}_{k} are denoted by , with f_{k-1,j}:=\mbox{\partial}_{k}(e_{k,j}) (Definition 3.3). Let
[TABLE]
By Assumption 4.9, we endow \mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x] with the . Using Definitions 5.2, where G_{0}=\mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}), with \mbox{\mathcal{B}}_{1} a basis of \mbox{\mathcal{C}}_{1} and \mbox{\partial}_{1}:\mbox{\mathcal{C}}_{1}\to\mbox{\mathcal{C}}_{0}, we endow \mbox{\mathcal{C}}_{1} with the monomial ordering induced by the one in \mbox{\mathcal{C}}_{0} and the subset . Taking into account that G_{k-1}=\mbox{\partial}_{k}(\mbox{\mathcal{B}}_{k}), with \mbox{\mathcal{B}}_{k} a basis of \mbox{\mathcal{C}}_{k} and \mbox{\partial}_{k}:\mbox{\mathcal{C}}_{k}\to\mbox{\mathcal{C}}_{k-1}, we can proceed recursively, and endow each \mbox{\mathcal{C}}_{k} with the monomial ordering induced by the one in \mbox{\mathcal{C}}_{k-1} and the subset .
Notation 6.4**.**
With the notations as above in Remark 6.3, set
[TABLE]
Similarly to Notation 5.4, are the monomial ideals of \mbox{\mathbb{A}}=\mbox{\mathbb{K}}[x] defined as follows: , and for each ,
[TABLE]
Note that, if \mbox{{\bf L}{\rm m}}(f_{k-1,j}) and \mbox{{\bf L}{\rm m}}(f_{k-1,i}) are two monomials of \mbox{\mathcal{C}}_{k-1} which are multiple of different basis elements, then \langle\mbox{{\bf L}{\rm t}}(f_{k-1,j})\rangle:\mbox{{\bf L}{\rm t}}(f_{k-1,i})=0. In such a case, we understand (see Definition 5.1)
Notation 6.5**.**
We will prove Theorem 6.1 by induction on the degree of the component of the complex. We specify now the statements to be shown.
Statement \mbox{{\bf H}}_{0}:
G_{0}=\mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a Gröbner basis of \ker(\mbox{\partial}_{0})\subset\mbox{\mathcal{C}}_{0}.
\mbox{{\bf L}{\rm t}}(\mbox{\partial}_{1}(C,\overline{C}))=(-1)^{0}x^{C\to\overline{C}}(C\cup\overline{C}), for each (C,\overline{C})\in\mbox{\mathcal{B}}_{1}.
Statement \mbox{{\bf H}}_{k}, for (recall that ):
G_{k}=\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) is a Gröbner basis of \ker(\mbox{\partial}_{k})\subset\mbox{\mathcal{C}}_{k} w.r.t. the monomial ordering defined as in Remark 6.3.
\mbox{{\bf L}{\rm t}}(\mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2}))=(-1)^{k}x^{I_{k+1}\to I_{k+2}}(I_{1},\ldots,I_{k+1}\cup I_{k+2}), for each in \mbox{\mathcal{B}}_{k+1}.
Statement \mbox{{\bf H}}_{n-1}: \mbox{\partial}_{n-1}:\mbox{\mathcal{C}}_{n-1}\to\mbox{\partial}_{n-2} is injective.
Remark 6.6**.**
We already know that \mbox{{\bf H}}_{0} holds. Indeed, by Corollary 4.18, G_{0}=\mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1}) is a Gröbner basis of {\rm Im}(\mbox{\partial}_{1})=I(\mbox{\mathcal{L}})=\ker(\mbox{\partial}_{0}). Moreover, by Lemma 4.10,
[TABLE]
where, according to Definition 3.3, is identified with the unit element of \mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x].
Remark 6.7**.**
If \mbox{{\bf H}}_{n-2} holds, then \mbox{{\bf H}}_{n-1} holds. Indeed, for each , the basis element e_{n-1,i}=(I_{1},\ldots,I_{n})\in\mbox{\mathcal{B}}_{n-1} is a cyclically ordered partition of into blocks, with . Thus, , where . Any different basis element e_{n-1,j}\in\mbox{\mathcal{B}}_{n-1}, with , will be of the same form: e_{n-1,j}=(\{j_{1}\},\ldots,\{j_{n-1}\},\{n\})\in\mbox{\mathcal{B}}_{n-1}. By part of \mbox{{\bf H}}_{n-2},
[TABLE]
and, similarly,
[TABLE]
One sees that, since and are different, then \mbox{{\bf L}{\rm t}}(f_{n-1,i}) and \mbox{{\bf L}{\rm t}}(f_{n-1},j) are multiple of different basis elements. Thus \langle\mbox{{\bf L}{\rm t}}(f_{n-1,j})\rangle:\mbox{{\bf L}{\rm t}}(f_{n-1,i})=0 and , for all (see Notation 6.4). By Theorem 5.5, we deduce that \ker(\mbox{\partial}_{n-1})=0.
Remark 6.8**.**
If \mbox{{\bf H}}_{k} holds, for every , then \mbox{\mathcal{C}}_{\mathbb{G}} is an exact complex. Indeed, by Remark 6.6 above, {\rm Im}(\mbox{\partial}_{1})=\ker(\mbox{\partial}_{0}). For , by part of \mbox{{\bf H}}_{k},
[TABLE]
Finally, by definition, \mbox{\partial}_{n}=0, so {\rm Im}(\mbox{\partial}_{n})=0. On the other hand, \mbox{\partial}_{n-1} is injective by \mbox{{\bf H}}_{n-1}.
Purpose 6.9**.**
In the light of Remarks 6.6, 6.7 and 6.8, the proof of the main theorem is reduced to show “\mbox{{\bf H}}_{k-1}\Rightarrow\mbox{{\bf H}}_{k}, for ”.
Before that, we need two important results. These are presented in the next two sections.
6.2. Identifying superfluous monomial generators of the
module quotients
Given a basis element e_{k,i}=(I_{1},\ldots,I_{k},I_{k+1})\in\mbox{\mathcal{B}}_{k}, we want to identify superfluous elements in , where (recall Notation 6.4 and Remark 5.6).
Lemma 6.10**.**
Fix an integer , with . Suppose that \mbox{{\bf H}}_{k-1} holds. Fix a basis element in \mbox{\mathcal{B}}_{k}. Let e_{k,j}=(J_{1},\ldots,J_{k},J_{k+1})\in\mbox{\mathcal{B}}_{k}, for some .
If for some , , then .
Suppose, on the contrary, that , so that .
Then,
[TABLE]
In particular, if (and so ), then
[TABLE]
If , let be such that . Then , so is superfluous.
Proof.
Note that is vacuous if . So, suppose momentarily, that and that , for some . By the hypothesis that \mbox{{\bf H}}_{k-1} holds, we have that
[TABLE]
and
[TABLE]
where . Then follows from the fact that the least common multiple of two monomial terms is zero provided the two basis elements involved are different (see Definition 5.1, ).
Suppose that , with . Note that this means that and are each partitions of the same set . Now the two basis elements involved in \mbox{{\bf L}{\rm m}}(\mbox{\partial}_{k}(e_{k,j})) and \mbox{{\bf L}{\rm m}}(\mbox{\partial}_{k}(e_{k,i})) are the same (see (58) above). Therefore, using Definition 5.1, , again:
[TABLE]
Now, using (41), we see that
[TABLE]
Thus
[TABLE]
Recalling (43) in Notation 4.12, we deduce:
[TABLE]
In particular, if , then and . Therefore
[TABLE]
This proves . Suppose now that , so and . Let be such that . Using , we have that:
[TABLE]
In particular, divides , which proves . ∎
The following is an easy remark but will prove to be very useful in organizing the basis elements of \mbox{\mathcal{B}}_{k}.
Remark 6.11**.**
Fix an integer , with . Let be a basis element of \mbox{\mathcal{B}}_{k}, where . Set
[TABLE]
The following conditions hold.
The set \mbox{\mathcal{B}}_{k,i} is empty if and only if . (For instance, for the first basis element
[TABLE]
of \mbox{\mathcal{B}}_{k}, considering the , we have \mbox{\mathcal{B}}_{k,1}=\varnothing.)
The cardinality of \mbox{\mathcal{B}}_{k,i} is \lvert\mbox{\mathcal{B}}_{k,i}\rvert=2^{\lvert I_{k+1}\rvert-1}-1. (This formula holds trivially if .)
If \mbox{\mathcal{B}}_{k,i}\neq\varnothing and in \mbox{\mathcal{B}}_{k,i}, then . Moreover \mbox{\mathcal{B}}_{k,j}\subsetneq\mbox{\mathcal{B}}_{k,i}.
If \mbox{\mathcal{B}}_{k,i}\neq\varnothing, let \varrho_{k,i}:\mbox{\mathcal{B}}_{k,i}\to\mbox{\mathcal{B}}_{k+1} be defined by
[TABLE]
while if \mbox{\mathcal{B}}_{k,i}=\varnothing, we set \varrho_{k,i}(\mbox{\mathcal{B}}_{k,i}):=\varnothing. Then is well-defined and injective.
If , then \varrho_{k,i}(\mbox{\mathcal{B}}_{k,i}) and \varrho_{k,j}(\mbox{\mathcal{B}}_{k,j}) are two disjoint sets.
Then \mbox{\mathcal{B}}_{k+1}=\bigcup_{i=2}^{r_{k}}\varrho_{k,i}(\mbox{\mathcal{B}}_{k,i}). In particular, r_{k+1}=\sum_{i=2}^{r_{k}}\lvert\mbox{\mathcal{B}}_{k,i}\rvert.
Proof.
Suppose that and let (I_{1},\ldots,I_{k-1},J_{k},J_{k+1})\in\mbox{\mathcal{B}}_{k}, with . Then
[TABLE]
Since , it follows that and so there does not exist any element in \mbox{\mathcal{B}}_{k,i}. On the other hand, if and with , just take and . Then is in \mbox{\mathcal{B}}_{k,i}. This proves .
Suppose that \mbox{\mathcal{B}}_{k,i}\neq\varnothing, i.e., . So . Then the elements of \mbox{\mathcal{B}}_{k,i} are in one-to-one correspondence with the non-empty subsets of the set . This proves .
Let e_{k,j}=(I_{1},\ldots,I_{k-1},J_{k},J_{k+1})\in\mbox{\mathcal{B}}_{k,i}, that is, . Then, according to the , precedes and so . Moreover, if is in \mbox{\mathcal{B}}_{k,j}, then , and so e_{k,h}\in\mbox{\mathcal{B}}_{k,i}. Furthermore, e_{k,j}\in\mbox{\mathcal{B}}_{k,i}, but e_{k,j}\not\in\mbox{\mathcal{B}}_{k,j}. This proves .
Clearly, if e_{k,j}=(I_{1},\ldots,I_{k-1},J_{k},J_{k+1})\in\mbox{\mathcal{B}}_{k,i}, then and . Thus is indeed in \mbox{\mathcal{B}}_{k+1}. Take and in \mbox{\mathcal{B}}_{k,i} such that . Then and, so , and , which proves .
Next we prove . Let and , with \mbox{\mathcal{B}}_{k,i}\neq\varnothing and \mbox{\mathcal{B}}_{k,j}\neq\varnothing, respectively. Suppose that is in \mbox{\mathcal{B}}_{k,i} and is in \mbox{\mathcal{B}}_{k,j}. Suppose also that . Then , for . In particular, . Thus . The rest follows by .
Finally, we prove that holds. Consider (H_{1},\ldots,H_{k+2})\in\mbox{\mathcal{B}}_{k+1}. Let us find two basis elements e_{k,i}=(I_{1},\ldots,I_{k+1})\in\mbox{\mathcal{B}}_{k} and in \mbox{\mathcal{B}}_{k,i}, so that , with , and such that
[TABLE]
This forces , for , and . Therefore must be equal to and . Thus, both
[TABLE]
are uniquely determined by (H_{1},\ldots,H_{k+2})\in\mbox{\mathcal{B}}_{k+1}. ∎
Having in mind Notation 6.4, Theorem 5.5 and Remark 5.7, then the preceding Lemma 6.10 and Remark 6.11 can be summarised as follows. Note that Lemma 6.10 discards elements which are known to be superfluous for certain. However, it does not ensure that we have discarded all of them. In fact, we will see that this is precisely characterised by the digraph being strongly complete (see Corollary 6.16).
Proposition 6.12**.**
Fix an integer , with . Suppose that \mbox{{\bf H}}_{k-1} holds. For each , then
[TABLE]
is a generating set for the module quotient . Let stand for a -syzygy associated to the -vector of f_{k-1,i}=\mbox{\partial}_{k}(e_{k,i}) and f_{k-1,j}=\mbox{\partial}_{k}(e_{k,j}), where e_{k,j}\in\mbox{\mathcal{B}}_{k,i}. Then there is a Gröbner basis of \ker(\mbox{\partial}_{k})\subset\mbox{\mathcal{C}}_{k} (w.r.t. the monomial ordering on \mbox{\mathcal{C}}_{k} induced by the monomial ordering on \mbox{\mathcal{C}}_{k-1} and the Gröbner basis ) of the form
[TABLE]
The cardinality of this Gröbner basis is \sum_{i=2}^{r_{k}}\lvert\mbox{\mathcal{B}}_{k,i}\rvert, which coincides with , the rank of \mbox{\mathcal{C}}_{k+1}.
Proof.
Indeed, Lemma 6.10 ensures that in order to find a generating set of we only need to consider the with e_{k,j}\in\mbox{\mathcal{B}}_{k,i}. By Theorem 5.5 and Remark 5.7, we deduce that indeed there is a Gröbner basis of \ker(\mbox{\partial}_{k}) of the form (59).
By Discussion 5.3, we know that if , then and . In particular, the union in (59) is of disjoint sets and
[TABLE]
Thus, the cardinality of this Gröbner basis is \sum_{i=2}^{r_{k}}\lvert\mbox{\mathcal{B}}_{k,i}\rvert. By Remark 6.11, , \sum_{i=2}^{r_{k}}\lvert\mbox{\mathcal{B}}_{k,i}\rvert=r_{k+1}. ∎
Remark 6.13**.**
With the notations above, if is strongly complete or, equivalently, if is a matrix, then
[TABLE]
For in the strongly complete case, the map \mbox{\mathcal{B}}_{k,i}\to\mbox{\mathcal{C}}_{k-1} defined by is injective. Indeed, if e_{k,j}\in\mbox{\mathcal{B}}_{k,i}, then (see (57) in Lemma 6.10), where and , with . Since , for all , it follows that determines . Hence , and so and hence are uniquely determined by .
6.3. The images of basis elements
are syzygies associated to S-vectors
Let us show that the differential \mbox{\partial}_{k+1} of the complex maps each basis element of \mbox{\mathcal{B}}_{k+1} to a -syzygy associated to an -vector, up to a sign (see Discussion 5.3 ).
Proposition 6.14**.**
Fix an integer , with . Suppose that \mbox{{\bf H}}_{k-1} holds. Fix a basis element in \mbox{\mathcal{B}}_{k+1}. Then
[TABLE]
where and . That is, -\mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2}) is a -syzygy associated to the -vector of f_{k-1,i}=\mbox{\partial}_{k}(e_{k,i}) and f_{k-1,j}=\mbox{\partial}_{k}(e_{k,j}).
Proof.
Note that, according to the , does indeed precede (see Remark 3.1) and it follows from the definition that e_{k,j}\in\mbox{\mathcal{B}}_{k,i} (see Remark 6.11). By Definition 3.3, we have:
[TABLE]
Set , for some , with . Moreover, set , for some . Then
[TABLE]
Since \mbox{\partial}_{k}\circ\mbox{\partial}_{k+1}=0, and \mbox{\partial}_{k}(e_{k,u})=f_{k-1,u}, for each , we obtain:
[TABLE]
Let us see that the left hand side of (61) coincides with the -vector of and .
Since \mbox{{\bf H}}_{k-1} holds, we have
[TABLE]
Similarly,
[TABLE]
Then
[TABLE]
[TABLE]
Therefore,
[TABLE]
Hence
[TABLE]
and, similarly, . Therefore, the -vector of and is equal to
[TABLE]
which is the left hand side of (61).
Now, let us prove that the right hand side of (61) is a standard expression for the -vector w.r.t. the Gröbner basis , with remainder zero. Note that is indeed a Gröbner basis of \ker(\mbox{\partial}_{k-1}) due to hypothesis \mbox{{\bf H}}_{k-1}. First, observe that the only coincident terms among the summands of
[TABLE]
are precisely the leading terms. Indeed:
[TABLE]
whereas
[TABLE]
Excluding the (underlined) coincident leading terms (note that hypothesis \mbox{{\bf H}}_{k-1} holds), the basis elements involved in (6.3) end either in or else in , while the basis elements involved in (6.3) end either in or else in . Therefore, the -vector of and cancels the leading terms of these expressions, but no other summand of is cancelled by any other summand of . In particular,
[TABLE]
Now, we need to identify the leading monomials of the summands on the right hand side of (61). For , since the hypothesis \mbox{{\bf H}}_{k-1} holds,
[TABLE]
Similarly, the leading monomial of the last summand on the right hand side of (61) is:
[TABLE]
All the monomials terms in (6.3) and in (6.3) appear in (6.3) and are different from the underlined leading term \mbox{{\bf L}{\rm t}}(m^{k}_{j,i}f_{k-1,i}). (By (41), .) Thus
[TABLE]
Concatenating (64) and (67), we deduce that the right hand side of (61) is indeed a standard expression for with remainder zero and w.r.t. the Gröbner basis .
Hence, the corresponding -syzygy associated to the -vector of the elements f_{k-1,i}=\mbox{\partial}_{k}(e_{k,i}) and f_{k-1,j}=\mbox{\partial}_{k}(e_{k,j}) is
[TABLE]
Comparing (6.3) with (6.3), we get \mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2})=-\tau_{k}(e_{k,i},e_{k,j}), the desired equality. ∎
As an immediate consequence of Proposition 6.14, we get the following result. The second part can be seen as a natural generalisation of Lemma 4.14.
Corollary 6.15**.**
Fix an integer , with . Suppose that \mbox{{\bf H}}_{k-1} holds. Then
\mbox{{\bf L}{\rm t}}(\mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+1},I_{k+2}))=(-1)^{k}x^{I_{k+1}\to I_{k+2}}(I_{1},\ldots,I_{k},I_{k+1}\cup I_{k+2}).
Moreover, the restriction of the map \mbox{\partial}_{k+1}:\mbox{\mathcal{C}}_{k+1}\to\mbox{\mathcal{C}}_{k} over the basis set \mbox{\mathcal{B}}_{k+1} is injective. In particular, since G_{k}:=\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}), then
[TABLE]
Proof.
By Proposition 6.14, and with the notations used there, and by (56) in Discussion 5.3,
[TABLE]
which proves .
Suppose that \mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2})=\mbox{\partial}_{k+1}(J_{1},\ldots,J_{k+2}). Hence their leading terms coincide. Using the first part , it follows that
[TABLE]
Therefore, , for every , and moreover, . Consider the summand
[TABLE]
of \mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2}), preceding the leading term and different from
[TABLE]
(see (3.3) in Definition 3.3 and Remark 3.4). On the one hand, this element coincides with
[TABLE]
On the other hand, this element must coincide with a summand of \mbox{\partial}_{k+1}(J_{1},\ldots,J_{k+2}). It follows that either and , or else, and . However, the second case cannot occur, because, since , it would follow that , a contradiction. Thus the first case holds, which clearly implies and . ∎
6.4. Proof of the main theorem
Now, we have all the ingredients to prove Theorem 6.1.
Proof.
Acording to Purpose 6.9, we have to show “\mbox{{\bf H}}_{k-1}\Rightarrow\mbox{{\bf H}}_{k}, for ”. So fix an integer , with and suppose that \mbox{{\bf H}}_{k-1} holds. Let us prove that \mbox{{\bf H}}_{k} holds. By Proposition 6.12, we know that there is a Gröbner basis , say, of \ker(\mbox{\partial}_{k}) of the form:
[TABLE]
This is a Gröbner basis of \ker(\mbox{\partial}_{k}) w.r.t. the monomial ordering on \mbox{\mathcal{C}}_{k} induced by the monomial ordering on \mbox{\mathcal{C}}_{k-1} and the Gröbner basis (see also Remark 6.3). Note that Proposition 6.12 ensures that the cardinality of is .
In Proposition 6.14, we have shown that, for any (I_{1},\ldots,I_{k+2})\in\mbox{\mathcal{B}}_{k+1},
[TABLE]
where and . That is, -\mbox{\partial}_{k+1}(I_{1},\ldots,I_{k+2}) is a -syzygy associated to the -vector of f_{k-1,i}=\mbox{\partial}_{k}(e_{k,i}) and f_{k-1,j}=\mbox{\partial}_{k}(e_{k,j}). In other words, any element of -G_{k}=-\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) is an element of . Observe that, by Corollary 6.15, , the cardinality of is also . Since the set is included in the Gröbner basis of \ker(\mbox{\partial}_{k}), and both and have the same cardinality, it follows that G_{k}=\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) is a Gröbner basis of \ker(\mbox{\partial}_{k}).
(To avoid the preceding discussion using cardinalities, we can argue as follows. Take , any element in : that is, a -syzygy associated to the -vector of f_{k-1,i}=\mbox{\partial}_{k}(e_{k,i}) and f_{k-1,j}=\mbox{\partial}_{k}(e_{k,j}), with e_{k,j}\in\mbox{\mathcal{B}}_{k,i}. Remark that, by Discussion 5.3, both basis elements and are uniquely determined by the syzygy . Write and , with . Consider the map \varrho_{k,i}:\mbox{\mathcal{B}}_{k,i}\to\mbox{\mathcal{B}}_{k+1} introduced in Remark 6.11. Take the element
[TABLE]
Observe that , for all , that and that . Thus and . Hence
[TABLE]
Then, by Proposition 6.14 again, \mbox{\partial}_{k+1}(\varrho_{k,i}(e_{k,j}))=-\tau_{k}(e_{k,i},e_{k,j}). That is, any element of is an element of -G_{k}=-\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}). In conclusion, G_{k}=\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) is a Gröbner basis of \ker(\mbox{\partial}_{k}).)
Since \mbox{{\bf H}}_{k-1} holds, it follows that part of \mbox{{\bf H}}_{k} also holds (recall Notation 6.5). Part of \mbox{{\bf H}}_{k} follows from Corollary 6.15, , and the hypothesis that \mbox{{\bf H}}_{k-1} holds. This completes the proof. ∎
We characterize now when \mbox{\mathcal{C}}_{\mathbb{G}} is a minimal resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}). This result generalises the first main result in [MaS], namely Theorem 2 there. In a follow-up paper, we shall investigate applications of our results, beginning with generalisations of the remaining results of [MaS], before branching out more widely.
Corollary 6.16**.**
Let be a strongly connected digraph or, equivalently, let be an matrix. Let \mbox{\mathcal{C}}_{\mathbb{G}} be its associated complex. Then
[TABLE]
is a minimal free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) if and only if is strongly complete.
Proof.
Suppose that is strongly complete. That \mbox{\mathcal{C}}_{\mathbb{G}} is a minimal free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}) follows from Theorem 6.1 and Remark 3.10.
Conversely, suppose that \mbox{\mathcal{C}}_{\mathbb{G}} is a minimal free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}). Consider with , together with a basis element of the form
[TABLE]
Then appears as a coefficient in the expression for \mbox{\partial}_{k}(\{i\},\{j\},I_{3},\ldots,I_{k+1}). By hypothesis, x^{a_{i,j}}_{i}\in\mbox{\mathfrak{m}}, and it follows that . Similarly, .
Next consider a basis element of the form (I_{1},\ldots,I_{k-1},\{i\},\{n\})\in\mbox{\mathcal{B}}_{k}. A similar argument shows that .
Finally, consider a basis element of the form (\{i\},I_{2},\ldots,I_{k},\{n\})\in\mbox{\mathcal{B}}_{k}. Again, a similar argument shows that , and the result follows. ∎
6.5. More on enumerations
Since \mbox{\partial}_{k+1}:\mbox{\mathcal{B}}_{k+1}\to\mbox{\mathcal{C}}_{k} is injective, it is natural to consider in \mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) the enumeration inherited by the enumeration of \mbox{\mathcal{B}}_{k+1} (see Corollary 6.15, ). This is not a minor matter as regards this paper. Indeed, to choose a “correct” enumeration in \mbox{\mathcal{B}}_{k}, and hence in G_{k-1}=\mbox{\partial}_{k}(\mbox{\mathcal{B}}_{k}), has been seen to be crucial in the proof of Lemma 6.10 and hence in the proof of the main theorem, Theorem 6.1. In the light of this comment, the next remark (and Remark 6.18 below) may help to clarify ideas.
Remark 6.17**.**
Let us consider in \mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1}) the enumeration inherited by \mbox{\mathcal{B}}_{k+1}. On the other hand, let us also consider in \Upsilon:=\bigcup_{i=2}^{r_{k}}\{\tau_{k}(e_{k,i},e_{k,j})\mid e_{k,j}\in\mbox{\mathcal{B}}_{k,i}\} the enumeration defined in Remark 5.8 (where is possibly larger than the set of syzygies considered in Remark 5.8). Then both enumerations agree, considering that -\mbox{\partial}_{k+1}(\mbox{\mathcal{B}}_{k+1})=\Upsilon (see the proof of Theorem 6.1).
Proof.
Let e_{k+1,l}=(I_{1},\ldots,I_{k+2})\in\mbox{\mathcal{B}}_{k+1} and \mbox{\partial}_{k+1}(e_{k+1,l})=-\tau_{k}(e_{k,i},e_{k,j}), with
[TABLE]
Analogously, let e_{k+1,h}=(H_{1},\ldots,H_{k+2})\in\mbox{\mathcal{B}}_{k+1} and \mbox{\partial}_{k+1}(e_{k+1,h})=-\tau_{k}(e_{k,p},e_{k,q}), with
[TABLE]
(See Proposition 6.14.) Suppose that \mbox{\partial}_{k+1}(e_{k+1,h}) precedes \mbox{\partial}_{k+1}(e_{k+1,l}) in the inherited enumeration by \mbox{\mathcal{B}}_{k+1}, that is, precedes . By Remark 3.1, there exists an integer , such that and precedes . If , then precedes . If , then and precedes . Therefore, precedes , according to the enumeration considered in Remark 5.8.
Conversely, suppose that precedes . If precedes , then there exists an integer , such that and precedes . This clearly implies that precedes . On the other hand, if and precedes , then, since , necessarily precedes . But, since , it follows that precedes and so precedes , and \mbox{\partial}_{k+1}(e_{k+1,h}) precedes \mbox{\partial}_{k+1}(e_{k+1,l}). ∎
Remark 6.18**.**
If we attempt to prove Theorem 6.1 by means of Theorem 5.5, only now using an alternative but seemingly natural enumeration of the basis elements of the successive free modules in the resolution, we can find that the module quotients have an ‘excessive’ number of generators, and as a result, the free resolution we obtain is no longer the complex.
For example, consider the matrix considered in [MaS, Example 1, 3]:
[TABLE]
Here . Thus each variable is given degree and we endow A=\mbox{\mathbb{K}}[x,y,z,t] with the (weighted) reverse lexicographic ordering (see Assumptions 2.7 and 4.9). Now, instead of enumerating according to the , we enumerate \mbox{{\rm Cyc}}_{4,2} as follows (see Example 3.2):
[TABLE]
Note that in Corollary 4.18, the enumeration considered in \mbox{\mathcal{B}}_{1} is not essential for its proof. Thus it follows that is a Gröbner basis of I(\mbox{\mathcal{L}}), where
[TABLE]
It is a simple check to see that ():
[TABLE]
That gives minimal generators of the ideal quotients , for , in Theorem 5.5. Thus the Gröbner basis of the syzygies of would contain elements and the corresponding resolution would begin with
[TABLE]
and so it would not coincide with the complex \mbox{\mathcal{C}}_{L}.
7. An example: the four-dimensional case
In this section we write down the case just as an illustrative example. This will enable the reader to visualise the details of the proofs of our main results for this particular case.
Recall that, when , the complex has the following form (cf. Example 3.8 for more details).
[TABLE]
where we write \mbox{\mathbb{K}}[x]\equiv\mbox{\mathbb{K}}[x,y,z,t]. Moreover, Theorem 6.1 says that this is a free resolution of \mbox{\mathbb{K}}[x]/I(\mbox{\mathcal{L}}). By Corollary 4.18, G_{0}:=\mbox{\partial}_{1}(\mbox{\mathcal{B}}_{1})\subset\mbox{\mathcal{C}}_{0} is a Gröbner basis of \mbox{\partial}_{1}(\mbox{\mathcal{C}}_{1})=I(\mbox{\mathcal{L}})=\ker(\mbox{\partial}_{0}). Using Definition 3.3 (see also Example 3.8), on enumerating according to the and underlining leading terms, the elements of are:
[TABLE]
Let us calculate the ideal quotients of Notation 6.4. To do that, we need to calculate m^{1}_{j,i}=\mbox{{\rm LCM}}(\mbox{{\bf L}{\rm m}}(f_{0,j}),\mbox{{\bf L}{\rm m}}(f_{0,i}))/\mbox{{\bf L}{\rm t}}(f_{0,i}), for . By definition . The subsequent is equal to . To calculate , observe that is a multiple of , so it follows that . Note that this is consistent with Lemma 6.10 and Proposition 6.12. Indeed, and
[TABLE]
Since , then e_{1,1}=(123,4)\in\mbox{\mathcal{B}}_{1,3} and, since , then e_{1,2}\not\in\mbox{\mathcal{B}}_{1,3}. In other words, is superfluous. The next are presented in the following tables:
[TABLE]
[TABLE]
Similarly, is generated by the monomials in the column labelled “”. But and are multiples of . Therefore, . Again, this is consistent with Proposition 6.12, since and are not in \mbox{\mathcal{B}}_{1,4}, so they are superfluous. As for , note that are in \mbox{\mathcal{B}}_{1,5}, but is not. Thus . Similarly, and . Observe that there are exactly twelve monomial generators, while \mbox{\mathcal{C}}_{2} is precisely the free module of rank , which is consistent with the equality r_{2}=\sum_{i=2}^{r_{1}}\lvert\mbox{\mathcal{B}}_{1,i}\rvert in Proposition 6.12. Concretely, the set of monomial generators of the ideal quotients are:
[TABLE]
Observe that all these are different provided that is a matrix, which is coherent with Remark 6.13. On the other hand, if is an matrix, which is not a matrix, then there may be coincidences (e.g., if and , then and are equal to ).
Each monomial generator of induces a corresponding -syzygy element associated to the -vector of and . Then Theorem 5.5 states that
[TABLE]
is a Gröbner basis of \ker(\mbox{\partial}_{1})\subset\mbox{\mathcal{C}}_{1} w.r.t. the monomial ordering on \mbox{\mathcal{C}}_{1} induced by the on \mbox{\mathcal{C}}_{0}=\mbox{\mathbb{K}}[x] and the Gröbner basis . Note that we have enumerated this Gröbner basis according to Remark 5.8. The proof of Theorem 6.1 and Remark 6.17 say that the following is an equality of enumerated sets:
[TABLE]
More concretely (recall Example 3.8), and underlining the leading terms (using either (56) or Theorem 6.1):
[TABLE]
Let us focus on the first equality. Since is a complex, \mbox{\partial}_{1}(f_{1,1})=\mbox{\partial}_{1}(\mbox{\partial}_{2}(e_{2,1}))=0. Hence
[TABLE]
In other words,
[TABLE]
On one hand, because and . On the other hand, (75) is a standard expression of w.r.t. the on \mbox{\mathcal{C}}_{0} and with respect to , and with remainder zero. Indeed, the -vector of and is defined so as to cancel the leading terms of and . Thus on the left hand side we have just a binomial, and hence similarly on the right hand side. Therefore, the leading term of the right hand side must necessarily be equal to the leading term of the left hand side. This proves that (75) is a standard expression. One deduces that is a -syzygy associated to the -vector of and . So f_{1,1}=\mbox{\partial}_{2}(e_{2,1})=-\tau_{1}(e_{1,2},e_{1,1}), which is consistent with Proposition 6.14 and Theorem 6.1.
The remainder of the equalities can be treated in the same manner, so we deduce that G_{1}=\mbox{\partial}_{2}(\mbox{\mathcal{B}}_{2}) is a Gröbner basis of \ker(\mbox{\partial}_{1}) and
[TABLE]
and the complex is exact in degree , as shown in Theorem 6.1.
We next find the syzygies of G_{1}=\mbox{\partial}_{2}(\mbox{\mathcal{B}}_{2})=\{f_{1,1},\ldots,f_{1,12}\}. To do that, we compute the corresponding module quotients of Notation 6.4. By definition, . Since the least common multiple of two monomial terms with different basis elements is zero, it follows that , and are zero. For the same reason, and . Note that is generated by . (Remark this negative sign, which will be the reason for the negative sign in the subsequent equality \mbox{\partial}_{3}(e_{3,1})=-\tau_{2}(e_{2,5},e_{2,4}).) Next, . Since e_{2,5}\not\in\mbox{\mathcal{B}}_{2,6}, then is superfluous. Thus . It is easy to check that the nonzero module quotients are:
[TABLE]
Observe that there are exactly six monomial generators, while \mbox{\mathcal{C}}_{3} is precisely the free module of rank , which is consistent with the equality r_{3}=\sum_{i=2}^{r_{2}}\lvert\mbox{\mathcal{B}}_{2,i}\rvert in Proposition 6.12. Concretely, the set of monomial generators of the ideal quotients are:
[TABLE]
Each monomial generator of produces a corresponding -syzygy element associated to the -vector of and . Then Theorem 5.5 affirms that
[TABLE]
is a Gröbner basis of \ker(\mbox{\partial}_{2})\subset\mbox{\mathcal{C}}_{2}. Again, we have enumerated this Gröbner basis according to Remark 5.8. Moreover, Theorem 6.1 and Remark 6.17 say that the following is an equality of enumerated sets:
[TABLE]
More concretely (recall Example 3.8), and underlining the leading terms (using either (56) or Theorem 6.1):
[TABLE]
[TABLE]
Again, let us focus on the first equality. Since is a complex, \mbox{\partial}_{2}(f_{2,1})=\mbox{\partial}_{2}(\mbox{\partial}_{3}(e_{3,1}))=0. Hence
[TABLE]
In other words,
[TABLE]
Note that , because and . On the other hand, (78) is a standard expression of with respect to , with remainder zero. Indeed, the -vector of and is defined so as to cancel the leading terms of and . Recall that
[TABLE]
Note that the leading term of and the leading term of coincide with the two summands of -x^{a_{1,4}}f_{1,5}-\mbox{{\bf L}{\rm t}}(-x^{a_{1,4}}f_{1,5}), as stated in the proof Proposition 6.14. It follows that (78) is a standard expression, so one deduces that
[TABLE]
is a -syzygy associated to the -vector of and . Thus, , which is consistent with Proposition 6.14 and Theorem 6.1.
The remainder of the equalities can be treated in the same manner, so we deduce that G_{2}=\mbox{\partial}_{3}(\mbox{\mathcal{B}}_{3}) is a Gröbner basis of \ker(\mbox{\partial}_{2}) and
[TABLE]
and the complex is exact in degree , as shown in Theorem 6.1.
Finally, observe that, since all the leading terms of have distinct basis elements, then the module quotients are all zero (see Notation 6.4). Therefore, there are no syzygies among the elements of and \mbox{\partial}_{3}:\mbox{\mathcal{C}}_{3}\to\mbox{\mathcal{C}}_{2} is injective, from which follows the exactness of the complex in degree , which is consistent with Remark 6.7.
Acknowledgements. In developing this research, we have benefitted from the hospitality of Departament de Matemàtiques, Universitat Politècnica de Catalunya; the School of Mathematics, University of Edinburgh; and ICMS, Edinburgh. We gratefully acknowledge financial support from the research grant MTM2015-69135-P and also grants from ICMS, Edinburgh, under their ‘Research in Groups’ scheme.
In particular, we would like to thank F.O. Schreyer for the stimulus provided by the several informative conversations we had at the June meeting on “Minimal free resolutions, Betti numbers, and combinatorics”, held at ICMS, Edinburgh.
We would also like to thank the referees for their helpful and interesting reports which have improved the present work.
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