Simplicial Resolutions of Powers of Square-free Monomial Ideals

The Taylor resolution is almost never minimal for powers of monomial ideals, even in the square-free case. In this paper we introduce a smaller resolution for each power of any square-free monomial ideal, which depends only on the number of generators of the ideal. More precisely, for every pair of fixed integers $r$ and $q$, we construct a simplicial complex that supports a free resolution of the $r$-th power of any square-free monomial ideal with $q$ generators. The resulting resolution is significantly smaller than the Taylor resolution, and is minimal for special cases. Considering the relations on the generators of a fixed ideal allows us to further shrink these resolutions. We also introduce a class of ideals called"extremal ideals", and show that the Betti numbers of powers of all square-free monomial ideals are bounded by Betti numbers of powers of extremal ideals. Our results lead to upper bounds on Betti numbers of powers of any square-free monomial ideal that greatly improve the binomial bounds offered by the Taylor resolution.


INTRODUCTION
Important insight about the underlying structure of an ideal in a polynomial ring is gained from a careful analysis of its minimal free resolution.As such, significant effort has gone into the development of methods to compute resolutions.The approach of leveraging connections between commutative algebra and other fields, such as combinatorics and topology, has proven to be quite fruitful.Diana Taylor's thesis [19] initiated the exploration of these connections, followed by simplicial resolutions (Bayer, Peeva, and Sturmfels [1]), polytopal complexes (Nagel and Reiner [15]), and cellular complexes (Bayer and Sturmfels [2]), to name just a few.See [14,16] for an overview of these developments.
The Taylor resolution is powerful: given any ideal I minimally generated by q monomials, Taylor constructed a simplicial complex Taylor(I) by labeling the vertices of a (q − 1)-simplex with the monomial generators of I.She showed that this complex supports a free resolution of I, in the sense that its simplicial chain complex can be transformed, via a process called homogenization, to a free resolution of I, called the Taylor resolution of I. Unfortunately, even though every monomial ideal has a Taylor resolution, the Taylor resolution is often far from minimal.In particular, for powers of ideals it is almost never minimal due to certain syzygies that are automatically created when taking powers.
The central theme of this paper is to find an analogue for the Taylor complex for powers of square-free monomial ideals.We seek a construction that takes the automatically generated non-minimal syzygies into account and removes them from the Taylor resolution to produce a much smaller free resolution of I r that works for any monomial ideal I. Our ultimate goal is to find a uniform combinatorial structure that depends only on the number of generators and the power of the ideal.More precisely, the question at the heart of this paper is the following: Question.Given positive integers r and q, is it possible to find a simplicial complex (considerably) smaller than the simplex Taylor(I r ) that supports a free resolution of I r , where I is any ideal generated by q monomials in a polynomial ring?
When r = 1, Taylor(I) is in fact the optimal answer to the question above, as there are ideals I for which Taylor(I) supports a minimal resolution.But when r = 2, the resolution supported on Taylor(I 2 ) is never minimal for any non-principal square-free monomial ideal I ( [4]).As expected, the resolution supported on Taylor(I r ) becomes further from minimal as r grows.
Although Taylor's complex for the r th power of a monomial ideal with q generators can be quite large -a simplex of dimension q+r−1 r -we can improve the situation considerably by studying the general relations among the generators of I r that must always exist for all monomial ideals regardless of the generating set for I. Similar to the idea used in Lyubeznik's resolutions [12], in the case of square-free monomial ideals, this investigation involves detecting and trimming redundant faces of the Taylor complex Taylor(I r ), bringing us closer to a minimal resolution.
To illustrate our underlying process, let r = 2, q = 3 and consider any ideal I = (m 1 , m 2 , m 3 ) in the polynomial ring k[x 1 , . . ., x n ] where m 1 , m 2 and m 3 are minimal, square-free monomial generators and k is a field.Now I 2 = (m 1 2 , m 2 2 , m 3 2 , m 1 m 2 , m 1 m 3 , m 2 m 3 ) and Taylor(I 2 ) is a 5-dimensional simplex with 6 vertices, where each vertex is labeled by a generator of I 2 and each face is labeled with the least common multiple of its vertices.A non-minimal syzygy occurs when a face and a subface have the same label (see Theorem 3.1).When considering I 2 , no matter what the monomial generators of I are, when i, j, k are distinct we always have the following: These equalities lead to non-minimal syzygies in the Taylor resolution of I 2 , and as a result the edges and all faces containing these edges, can be removed from Taylor(I 2 ).The resulting 2-dimensional subcomplex L 2 3 of Taylor(I 2 ) supports a resolution of I 2 ( [4]).This simple observation has a considerable impact on bounding the Betti numbers of I 2 .For example, in this small case we can conclude that for every ideal I with three square-free monomial generators, the projective dimension of I 2 is at most 2 (the dimension of L 2 3 ), versus the bound 5 (the dimension of Taylor(I 2 )).
In this paper we show that a similar argument can be made more generally.If I is minimally generated by square-free monomials m 1 , . . ., m q , then we can identify faces of the simplex Taylor(I r ) which lead to non-minimal syzygies for I r .Eliminating these faces results in a much smaller subcomplex of Taylor(I r ), which we call L r q (Definition 4.2).Our main result, Theorem 5.9, shows that L r q supports a resolution of I r .This echoes the power of Taylor's complex in that we have a topological structure depending solely on r and q that supports a resolution of I r for any square-free monomial ideal I with q minimal generators.By further deleting redundancies specific to the generators of I, we obtain a subcomplex L r (I) of L r q , so that we have L r (I) ⊆ L r q ⊆ Taylor(I r ).Theorem 5.9 also shows that L r (I) supports a free resolution of I r .Our approach to proving this involves showing that L r (I) and L r q are both quasi-trees (see Definition 2.2), meaning (roughly) that are built from a special ordering on their facets.In Theorem 3.6, we show that for a quasi-tree to support a free resolution one only needs to check that certain induced subcomplexes are connected.This extends the main result of [9].
Our complex L r (I) gives natural and useful bounds on homological information of I r .Indeed, the Betti numbers of I r are bounded in terms of the number and size of faces of L r (I), yielding bounds that are significantly smaller than those given by the Taylor resolution.In the later sections of this paper we examine just how much smaller these bounds on the Betti numbers are and when the resolutions obtained are minimal.We define a class of square-free monomial ideals, which we call extremal ideals, whose Betti numbers of powers bound the Betti numbers of powers of any square-free monomial ideal with the same number of generators.As a result, we reduce the question of minimality of L r q to the study of when L r q supports a minimal free resolution of the r th power of an extremal ideal.
To put this work in context of the broader literature, studying powers of ideals and bounding their invariants has received much attention in recent years.Powers play an important role in Rees algebras and associated graded rings among other uses, making understanding their behavior desirable but difficult.In another direction, there has been considerable interest in describing minimal topological resolutions for all monomial ideals using a variety of methods, such as using chain maps from multiple simplicial complexes (see [7,20]).In this paper we combine the two interests and seek, for powers of monomial ideals, resolutions that are supported on a single topological structure which is practical to determine based on the generators of the original ideal I.
The paper is organized as follows.Section 2 contains basics of simplicial resolutions.In Section 3 we use simplicial collapsing and the Bayer-Peeva-Sturmfels criterion to prove the above-mentioned criterion for quasi-trees (Theorem 3.6).In Section 4 we introduce the definition of the simplicial complex L r q , and we prove in Proposition 4.6 that L r q is a quasi-tree.In Section 5 we define L r (I), discuss some examples, and then prove the main result, Theorem 5.9.Section 6 investigates the bounds on the Betti numbers of I r that follow from the main result.Finally, Section 7 introduces extremal ideals, which have maximal Betti numbers among powers of square-free monomial ideals.In particular, Proposition 7.11 provides a full characterization of the conditions on r and q that guarantee L r q supports a minimal free resolution of I r for some ideal I.This paper is an extension of the work in [4] where the focus is on the second power of the ideal I.The collaboration was initiated at the 2019 workshop "Women in Commutative Algebra" hosted by the Banff International Research Station.

SIMPLICIAL RESOLUTIONS
Fix S = k[x 1 , . . ., x n ] to be a polynomial ring over a field k.We begin by reviewing necessary background for simplicial complexes and simplicial resolutions and then demonstrate the potential relationship to resolutions of ideals.
A simplicial complex ∆ on a vertex set V is a set of subsets of V such that if F ∈ ∆ and G ⊆ F then G ∈ ∆.We use the following terminology for simplicial complexes: Definition 2.1.Let ∆ be a simplicial complex.
(1) An element of ∆ is called a face.
(2) The facets of ∆ are the maximal faces under inclusion.
(3) The dimension of a face The dimension of ∆ is the maximum of the dimensions of its faces.
(5) ∆ is called a simplex if it has one facet.(6) The f -vector f (∆) = (f 0 , . . ., f d ) of a d-dimensional simplicial complex ∆ has f i = the number of i-dimensional faces of ∆.
Note that a simplicial complex can be uniquely determined by its facets.One writes ∆ = F 1 , . . ., F q to denote a simplicial complex ∆ with facets F 1 , . . ., F q .
In subsequent sections, we will use the tool of trimming simplicial complexes via certain rules.Essentially, we will delete vertices in a specified fashion.Vertex deletions naturally lead to the consideration of subcomplexes of simplicial complexes, which are defined below, together with additional structures and notions that will be used throughout the paper.Definition 2.2.Let ∆ be a simplicial complex on a vertex set V .
(1) If v is a vertex of ∆, then the deletion of v from ∆ is the simplicial complex (2) A subcomplex of ∆ is a subset of ∆ which is also a simplicial complex.
(3) Given W ⊆ V , the induced subcomplex of ∆ on W is the subcomplex for all facets H = F .The facet G is called a joint of F .(Note that the joint of a leaf need not be unique (see [8]). ( 5) ∆ is called a quasi-forest [23] if the facets of ∆ can be ordered as F 1 , . . ., F q such that for i = 1, . . ., q, the facet F i is a leaf of the simplicial complex F 1 , . . ., F i .(6) ∆ is a quasi-tree if it is a connected quasi-forest.
Example 2.3.The simplicial complex below is a quasi-tree.The leaf order is F 1 , . . ., F 5 , meaning that each F i is a leaf of F 1 , . . ., F i .In this example the joint of F i is F i−1 for all i ≥ 1.
Example 2.4.The star-shaped complex drawn below on the left is a quasi-tree, with leaf order F 0 , F 1 , F 2 , F 3 .
In particular, the center facet F 0 is the joint of F i for every i ≥ 1.This complex is a standard example of a quasi-tree which is not a simplicial tree in the sense of [8].This particular quasi-tree is shown in [4] to support a free resolution of the second power of any ideal with three square-free monomial generators.
If one removes F 0 from the center, the remaining complex is shown in the picture on the right.This simplicial complex is not a quasi-tree, since no facet is a leaf.
quasi-tree not a quasi-tree A free resolution of I is an exact sequence of the form where S βj is a free S-module of rank β j and t ∈ N. When β j is the smallest possible rank of a free module in the j th spot of any free resolution of I for each j, the resolution is minimal.In this case, the numbers β j are invariants of I and are called the Betti numbers of I.
In the 1960s, Diana Taylor demonstrated a striking connection between a (q − 1)-simplex and a resolution of a monomial ideal I in S. If I is a monomial ideal in S minimally generated by monomials m 1 , . . ., m q , then Taylor(I) denotes the simplex with q vertices indexed by the set [q] = {1, . . ., q}, where each vertex i is labeled with one of the monomials m i , and each face σ is labeled with the monomial In her Ph.D thesis [19], Taylor proved that the simplicial chain complex of Taylor(I) gives rise to a multigraded free resolution of I.In particular, the i th Betti number of I is bounded above by the number of idimensional faces of Taylor(I), which is q i+1 .This method has been generalized so that if ∆ is a simplicial or cellular complex whose vertices are labeled with the monomial generators m 1 , . . ., m q of an ideal I and whose faces are labeled with the least common multiple of the vertex labels as above, then we say that ∆ supports a free resolution of I if the homogenization of the simplicial (or cellular) chain complex of ∆ is a multi-graded free resolution of I, denoted by F ∆ , see [1,2].The multi-graded complex F ∆ is described as follows.For each t ≥ 0, the free module (F ∆ ) t has basis elements denoted by e σ , where σ ranges over all faces of ∆ with |σ| = t + 1, and e σ is considered to have multi-degree M σ .The differential is described by where Example 2.5.Let I = (xy, yz, zu) in R = k[x, y, z, u].The labeled simplicial chain complex supports a free resolution of I.The chain complex of ∆ is and homogenization results in the free resolution where the notation R(x a y b z c u d ) refers to the R-free module with one generator in multi-degree (a, b, c, d).
Remark 2.6.Simplicial complexes that support a free resolution of a monomial ideal are usually constructed such that the vertices correspond to and are labeled by a minimal set of generators of the ideal.However, one can also work with non-minimal generators, at the expense of producing a larger complex.In particular, one can mimic the construction of Taylor(I), but use instead any set of monomial generators of I.The same considerations show that this complex supports a free resolution of I.
There are various combinatorial ways to build subcomplexes ∆ of Taylor(I) that support a free resolution of I.One such well known complex is the Lyubeznik complex, which supports the Lyubeznik resolution of I ( [12]).The Lyubeznik resolution is the main inspiration for the complexes L r q and L r (I) which appear later in this paper.We will not define this resolution since it is not used in this paper, but refer the reader to [12,13] for additional information.

RESOLUTIONS SUPPORTED ON QUASI-TREES
Taylor's resolution is usually far from minimal.However, for a given monomial ideal I, a criterion of Bayer, Peeva and Sturmfels (see Theorem 3.1) allows one to check if a subcomplex of Taylor(I) supports a free resolution of I.In this section, using the above criterion and by observing that quasi-trees are collapsible, we show that a quasi-tree ∆ supports a free resolution of a given monomial ideal if and only if certain subcomplexes of ∆ are connected.
For a subcomplex ∆ of Taylor(I) and a monomial M in S, let ∆ M be the subcomplex of ∆ induced on the vertices of ∆ whose labels divide M .
The following is the criterion of Bayer, Peeva and Sturmfels [1, Lemma 2.2]; see also [6, Theorem 2.2] for the statement on minimality.
Theorem 3.1 (Criterion for a simplicial complex to support a free resolution).Let ∆ be a simplicial complex whose vertices are labeled with a monomial generating set of a monomial ideal I in a polynomial ring S over a field.Then ∆ supports a free resolution of I over S if and only if for every monomial M , the induced subcomplex ∆ M of ∆ on the vertices whose labels divide M is empty or acyclic.
Furthermore, a free resolution supported on ∆ is minimal if and only if M σ = M σ ′ for every proper subface σ ′ of a face σ of ∆.
Remark 3.2.The results of Theorem 3.1 are usually stated with the assumption that one uses the minimal monomial generating set of I for the labels.However, the proof of [1, Lemma 2.2] does not make use of this assumption, hence we formulated the result above to reflect this observation.
In particular, Theorem 3.1 implies that the f -vector of a complex ∆ supporting a resolution of a monomial ideal I is an upper bound for the vector of Betti numbers of I.In other words, for each i ≤ d = dim(∆), In particular, if ∆ supports a minimal free resolution of I, then equality holds above.
Using Theorem 3.1 it is straightforward to see that to determine whether ∆ supports a free resolution of I, it suffices to check that ∆ M is empty or acyclic only for monomials M in the lcm lattice of I; that is, for monomials M that are least common multiples of sets of vertex labels.
If the complex ∆ under consideration in Theorem 3.1 is a simplicial tree, then it suffices to show that ∆ M is connected, instead of acyclic, see [9].More precisely, it is established in [9] that every induced subcomplex of a simplicial tree is a simplicial forest, and then it is shown that simplicial trees are acyclic, and hence an induced subcomplex of ∆ is acyclic if and only if it is empty or connected (see [9, Theorems 2.5, 2.9, 3.2]).
We now generalize the work in [9] by showing that the criterion in Theorem 3.1 can be extended to the class of quasi-trees.To do so we need to argue that quasi-trees, and their connected induced subcomplexes, are acyclic.We do so using the following series of results.
Proof.By [10, Proposition 6], a simplicial complex ∆ with vertex set V is a quasi-forest if and only if for every subset W ⊆ V , the induced subcomplex ∆ W has a leaf.If W ⊆ V , consider the induced subcomplex ∆ W of ∆.If U ⊆ W , then ∆ U = (∆ W ) U has a leaf, and hence, ∆ W is a quasi-forest.
A face σ of a simplicial complex ∆ is called a free face if it is properly contained in a unique facet F of ∆.A collapse of ∆ along the free face σ is the simplicial complex obtained by removing the faces τ such that σ ⊆ τ ⊆ F from ∆.If additionally dim(σ) = dim(F ) − 1, then the collapse is called an elementary collapse.The simplicial complex ∆ is called collapsible if it can be reduced to a point via a series of (elementary) collapses.
Example 3.4.Consider the quasi-tree in Example 2.4.To illustrate that this complex is collapsible to a point, we label in Figure 1 the vertices and then demonstrate one collapsing sequence.At each step, the two faces that play the roles of σ and F in the exposition above are indicated.Note that the first two steps are elementary collapses for demonstration purposes.Alternate collapsing sequences exist.
In general, by starting with a leaf and the face that contains it, any quasi-tree can be collapsed to a point.

Proposition 3.5. Every quasi-tree is collapsible.
Proof.Proceed by induction on q, the number of facets of ∆.When q = 1, ∆ is a simplex, and known to be collapsible (e.g.[9,Proposition 2.7]).Suppose the statement holds for all quasi-trees with fewer than q facets.
Let ∆ be a quasi-tree with facet ordering F 1 , . . ., F q so that, F q collapses down to the face F ′ by removing faces outside ∆ ′ .As a result, this series of collapses brings ∆ to the quasi-tree ∆ ′ , which, by the induction hypothesis, is collapsible.
We now use Proposition 3.5 and Proposition 3.3 to show that the Bayer-Peeva-Sturmfels criterion in Theorem 3.1 can be extended to the class of quasi-trees.Theorem 3.6 (Criterion for a quasi-tree to support a free resolution).Let ∆ be a quasi-tree whose vertices are labeled with a monomial generating set of a monomial ideal I in a polynomial ring S over a field.Then ∆ supports a free resolution of I over S if and only if for every monomial M , the induced subcomplex ∆ M of ∆ on the vertices whose labels divide M is empty or connected.Proof.By Theorem 3.1, ∆ supports a resolution of I if and only if ∆ M is empty or acyclic for every monomial M ∈ S. If M ∈ S is a monomial, then by Proposition 3.3 above, ∆ M is a quasi-forest.Moreover, by Proposition 3.5, every connected component of ∆ M is contractible, i.e., homotopy equivalent to a point, and hence acyclic.The only possible homology that ∆ M could have would be that which comes from it being disconnected.As a result, ∆ supports a resolution of I if and only if ∆ M is empty or connected for every monomial M ∈ S.

THE QUASI-TREE L r q
Recall that our goal is to find a simplicial complex smaller than Taylor's complex that supports a free resolution of I r when I has q square-free monomial generators.In this section we construct a subcomplex L r q of the Taylor simplex, which depends only on the integers r and q, and contains a subcomplex supporting a free resolution of I r .
The base case for our construction is the case r = 1.In this case L 1 q is the well-known Taylor complex [19]: a simplex with q vertices that supports a free resolution of any monomial ideal with q generators.The case r = 2 was investigated in the earlier work of the authors [4].For example, it was shown in [4] that L 2 3 is the quasi-tree in Example 2.4.
We now collect and formalize the notation needed for the extension to the case r ≥ 3.
Notation 4.1.Let r and q be two positive integers.
• Let e 1 , . . ., e q denote the standard basis vectors for R q ; i.e., for each i ∈ [q], e i is the q-tuple with 1 in the i th coordinate and 0 elsewhere.
• Define N r q to be the set of points in Z q ≥0 whose coordinates add up to r: • Set s = r 2 ; that is, when r is odd, r = 2s − 1, and when r is even, r = 2s.Definition 4.2 (The simplicial complex L r q -see also Proposition 4.3).Let r, q ≥ 1 be two integers.Following Notation 4.1 we define L r q to be the simplicial complex with vertex set N r q whose faces are all subsets of the (not necessarily distinct) sets F r 1 , . . ., F r q , G r 1 , . . ., G r q defined as F r i = {(a 1 , . . ., a q ) ∈ N r q : a i ≤ max{r − 1, s} and a j ≤ s for j = i} . We refer to the set {F r 1 , . . ., F r q } as the first layer of L r q and the set {G r 1 , . . ., G r q } as the second layer of L r q .We define the base of L r q to be the face B r = (a 1 , . . ., a q ) ∈ N r q : a i ≤ s for all i ∈ [q] , so that F r i = B r ∪ {(a 1 , . . ., a q ) ∈ N r q : s + 1 ≤ a i ≤ r − 1}.In general, F r 1 , . . .F r q , G r 1 , . . ., G r q are the facets of L r q (Proposition 4.3); however, for small values of r and q, these sets need not be distinct.We summarize these facts in Proposition 4.3 to give a more precise description of the facets of L r q .Proposition 4.3 (Equivalent definition of L r q ).The simplicial complex L r q in Definition 4.2 can be described in terms of its distinct facets: Proof.When r = 1, then s = 1 and N 1 q = {e 1 , . . ., e q }.It follows that In this case L 1 q is a simplex with q vertices (the Taylor simplex).If q = 1 and r > 1, then When q ≥ 2 and r = 3, or q > 2 and r = 2, then the G r i are distinct as each G r i is the unique facet containing the vertex re i .Furthermore, since s + 1 = r, r contains all the vertices (r − 1)e i + e j where i = j, and so B r is not embedded in any of the G r i .Finally, when q ≥ 2 and r > 3, then s < r − 1.Therefore, the G r i are distinct, since each G r i is the unique facet containing the vertex re i .
For i, j The geometric realization of the simplicial complex L 3  2 is a path of length 3, as can be seen in the figure below.
For instance, if r = 6, then L 6  2 is pictured below, and B 6 is the single point at the middle of the 'bow-tie' and only the F r i s and G r j s are facets.
Note that the points in N r q can be viewed as lattice points in R q .Indeed, they are precisely the integer lattice points in a hyperplane section of the first octant, cut out by the hyperplane whose equation is While for small values of q this gives a natural way to depict the points, it does not illustrate the simplicial structure well.For instance, in Example 4.4, the 6 points would be co-linear, while our depiction emphasizes the existence of the two triangles.Using the combinatorial approach rather than the embedding in R q also allows for a generalized depiction, seen in the following example, that can easily be extended to higher q and r.
Example 4.5.We examine the case q = 3 in the same manner as above.The simplicial complexes L 1 3 , L 2 3 and L 3 3 appear in Figure 2, and L 6 3 appears in Figure 3.
The following statement generalizes [4, Proposition 3.3], which deals with the special case r = 2.
Proposition 4.6 (L r q is a quasi-tree).The simplicial complex L r q is a quasi-tree.
Proof.Following Proposition 4.3 for a description of the facets of L r q , we consider each case separately.When r = 1 or q = 1, L 1  1 is a simplex, and hence by definition a quasi-tree.When r = q = 2, L 2 2 has only two facets, and it is trivially a quasi-tree.
When r = 2 and q > 2, then order the facets of L 2 q by B 2 , G 2 1 , . . ., G 2 q .In this case, if i = j with i, j ∈ [q], then i with joint B 2 .When r = 3 and q ≥ 2, then as in the previous case, When r > 3 and q ≥ 2, we claim is a leaf order for L r q , making it a quasi-tree.To see this note that if i = j, i, j ∈ [q], then with joint F r j , and for each j ∈ {2, . . ., q} F r j is a leaf of F r 1 , . . ., F r j with joint, say, F r 1 .Hence (4.6.1) is a leaf order and L r q is a quasi-tree.

THE QUASI-TREE L r (I) SUPPORTING A FREE RESOLUTION OF I r
Given an ideal I with q square-free monomial generators, we now define an induced subcomplex of L r q , denoted L r (I), which is obtained by deleting vertices representing redundant generators of I r .We show in Theorem 5.9 that both L r q and L r (I) support a free resolution of I r .Definition 5.1 (The simplicial complex L r (I)).Let I be an ideal minimally generated by monomials m 1 , . . ., m q in the polynomial ring S. For a = (a 1 , . . ., a q ) ∈ N r q we set We use these equivalence classes to build an induced subcomplex L r (I), of L r q using the following steps: Step 1. From each equivalence class V i pick a unique representative c i in the following way: if Step 2. From the set {c 1 , . . ., c t }, eliminate all c i for which m cj | m ci for some j = i.We call the remaining set, without loss of generality, {c 1 , . . ., c u }.
Define L r (I) to be the induced subcomplex of L r q on the vertex set V .
The complex L r (I) is a subcomplex of Taylor(I r ).While L r (I) is dependent on the choices made when building its vertex set V , we will abuse the notation and ignore these choices, as they do not have an impact on our results.Also note that the set of monomials {m c1 , . . ., m cu } forms a minimal monomial generating set for the ideal I r .There are known classes of ideals, for instance, square-free monomial ideals of projective dimension one [3, Proposition 4.1], for which the generating set {m a | a ∈ N r q } does not contain redundancies, in which case L r q = L r (I).However, in general, L r (I) will be a proper subcomplex of L r q .Information about known redundancies in {m a | a ∈ N r q } can be used to produce L r (I).As an example, consider edge ideals of graphs.For such ideals, the redundancies in {m a | a ∈ N r q } are encoded in the ideal of equations of the fiber cone of the ideal, whose generators correspond to primitive even closed walks in the graph (see [22] for relevant definitions and details).One can check that the above relations are the only ones.Since r = 3, we have s = 2 and so all the above duplicated vertices are in B 3 .In particular, L 3 (I) will be the induced subcomplex of L 3 4 on a vertex set V with 16 vertices.The sets below are two different possible sets of vertices V for L 3 (I):  For r = 6 we have s = 3, and so 4e 1 + e 2 + e 3 ∈ F 6 1 \ B 6 but e 4 + e 5 + e 6 + e 7 + e 8 + e 9 ∈ B 6 .Therefore, the induced subcomplex L 6 (I) would not contain the vertex 4e 1 + e 2 + e 3 corresponding to m 1 4 m 2 m 3 .
While the examples above show that, in general, L r (I) is smaller than L r q , there are also cases when the two complexes are the same.In Section 7 we identify an ideal E q with L r (E q ) = L r q for each q.The two complexes are also equal for all I with q ≤ 3, as shown below.
Proposition 5.4.Let I be an ideal minimally generated by square-free monomials m 1 , . . ., m q in the polynomial ring S. If q ≤ 3 then L r (I) = L r q for all r ≥ 1. Proof.If q = 1, then I r = (m 1 r ) for all r, and both L r (I) and L r q consist of a single point, so the equality holds.
Assume 1 < q ≤ 3.By Definition 5.1 it suffices to show that L r (I) and L r q have the same vertex set, or in other words If a = b, then we may assume a 1 > b 1 and a 2 < b 2 .Since the monomials m i are square-free, we have which contradicts the minimality of the generators of I. Assume a = b.If a i = b i for some i then we can reduce to the case q = 2, so we may assume a i = b i for all i.Without loss of generality, assume a 1 > b 1 .
We have three cases: (1) Suppose a 2 > b 2 .In this case, we must also have a 3 < b 3 .Then Since m 1 and m 3 are square-free, this implies that m 1 | m 3 which gives a contradiction.(2) Suppose a 2 < b 2 and a 3 < b 3 .We have A similar contradiction is obtained when x ∤ m 2 .We conclude that for every x | m 1 , we must have x | m 2 and x | m 3 .This implies m 1 | m 2 and m 1 | m 3 , which is a contradiction.
Remark 5.5.Note that the square-free assumption is necessary in Proposition 5.4, for if I = (x 2 y, yz 2 , xyz), then m 1 m 2 | m 3 2 in I 2 , but (1, 1, 0) = (0, 0, 2) in N 2 3 .The next proposition shows that the vertices labeled by re i belong to the induced subcomplex L r (I) for all i ∈ [q], regardless of the choices made in Definition 5.1.Moreover, if q ≥ 2, then for each i ∈ [q] there exists some j ∈ [q] \ {i} such that the vertex labeled by (r − 1)e i + e j belongs to L r (I).
In what follows we use the standard notation LCM(I r ) to denote the lcm lattice of I r , which is the set of all least common multiples of subsets of the minimal monomial generating set of I r , partially ordered by division.
Proposition 5.6.Let r ≥ 1, I an ideal in S minimally generated by square-free monomials m 1 , . . ., m q , and i In particular, for all i, and all j as in (ii), Proof.We first observe that, for g, h > 0 and u, v ∈ [q], we have: Let a = (a 1 , . . ., a q ) ∈ N r q , and suppose m a | m i r .If, for some j ∈ [q], a j = 0 then m j aj | m i r , which by (5.6.1)implies that j = i, which results in a = re i .Thus (i) holds.
We now prove (ii).If r = 1 the statement is trivial, so assume r, q ≥ 2 and M ∈ LCM(I r ) satisfies Choose j ∈ A {i} such that m j has the fewest number of variables not dividing m i .Now suppose for some a = (a 1 , . . ., a q ) ∈ N r q , If a j ≥ 2, by canceling a copy of m j in (5.6.2) we will have m aj −1 j | m i r−1 which by (5.6.1)implies that i = j, a contradiction.Similarly if a i = r then by canceling a copy of m i r−1 in (5.6.2) we obtain m i | m j , again a contradiction since m i and m j are minimal generators.So we must have a j ≤ 1 and a i ≤ r − 1.
Now we claim that we must have a j = 1 or a i = r − 1.Otherwise, if a j = 0 and a i ≤ r − 2, since a 1 + • • • + a q = r, there must exist c, d ∈ [q] {i, j} with a c , a d > 0 (c and d could be equal).In particular by (5.6.2) (5.6.4) From (5.6.4) one can see that m ′ c and m ′ d do not share any variables with m i , and so by (5.6.3) and also (5.6.6)On the other hand m i | M , so (5.6.4) and ( 5 The fact that j ∈ A was picked so that m j has the fewest number of variables outside m i and (5.6.5)together imply d and m ′ j all have positive degrees.So the only possibilities are a j = 1 or a i = r − 1.If a j = 1 then by (i) we have m a = m i r−1 m j , and we are done.If , hence by cancellation k = j and m a = m i r−1 m j .The remaining statements now follow immediately from (i), (ii) and Definition 5.1.
One additional lemma is necessary before the statement of our main result.
Lemma 5.7.Using Notation 4.1, let i, j ∈ [q] with i = j, and let a = (a 1 , . . ., a q ) and b = (b 1 , . . ., b q ) be in N r q .If α is a non-negative integer such that a i ≥ α and b j > r − α Let x be a variable such that x | m j .Then x bj | m b and hence x bj | m a .Suppose x does not divide m i .Then x bj | m a ′ .Since m a ′ is a product of r − a i square-free monomials, we have then The hypothesis on a i and b j gives r − a i ≤ r − α < b j .The contradiction that arises shows x | m i .We conclude m j | m i , hence i = j, a contradiction.Now we prove (ii).Set It will be shown that for all variables x, where for a monomial M , deg x (M ) = max{t : Since m a ′ is a product of r − a i square-free monomials, and m b ′ is a product of r − b j square-free monomials, we have 0 ≤ u ≤ r − a i and 0 ≤ v ≤ r − b j . (5.7.1) This finishes the proof of (ii).
Finally, (iii) follows directly from the fact that Remark 5.8.Recall that s = r 2 .When α = s, the condition on b j in Lemma 5.7 can be translated as follows: Recall that, by Theorem 3.6, if ∆ is a quasi-tree with vertices labeled with the monomial generating set of a monomial ideal I ⊆ k[x 1 , . . ., x n ], then ∆ supports a resolution of I if and only if ∆ M is connected for every monomial M .It is easy to check that the connectivity need only be verified for M in LCM(I).We now use this criterion to prove the main result of the paper.Theorem 5.9 (Main Result).Let q ≥ 1 and let I be a monomial ideal minimally generated by square-free monomials m 1 , . . ., m q .By labeling each vertex a of the simplicial complexes below with the monomial m a , the following hold for any r ≥ 1: (1) L r q supports a free resolution of I r ; (2) L r (I) supports a free resolution of I r .
Proof.In this proof, we let L denote either L r q or L r (I).Fix L as one of these two choices.Let V denote the set of vertices of L and let {m 1 , . . ., m q } be the minimal monomial generating set of I. Following Notation 4.1, for a = (a 1 , . . ., a q ) ∈ N r q we let m a = m 1 a1 • • • m q aq and e 1 , . . ., e q denote the standard basis vectors for R q .
We willl show that for every M in LCM(I r ), L M is empty or connected, where L M is the induced subcomplex of the complex L on the set By Proposition 4.6, L r q is a quasi-tree and by Proposition 3.3, L r (I) is a quasi-forest, which is connected and thus a quasi-tree.In view of Theorem 3.6, we can then conclude that L supports a free resolution of I r .
Suppose M ∈ LCM(I r ) and L M = ∅.If M = m i r for some i ∈ [q], then L M is a point by Proposition 5.6 (i), and hence is connected.
Assume M = m i r for all i ∈ [q].Note that this implies q > 1.The facets of L M are the maximal sets, under inclusion, among the sets (5.9.1) Note that not all these sets are facets of L M , but all facets of L M are among those listed in (5.9.1).We refer to the facets of L M of form F r i ∩ V M as the first layer, and those of the form G r i ∩ V M as the second layer.We refer to B r ∩ V M as the base of L M .The base B r ∩ V M could become empty, depending on the choice of M .
We use the faces in (5.9.1) to argue the connectedness of L M as follows: Claim 1 below shows that any facet of L M in the second layer is connected to a facet in the first layer.Claim 2 implies that any two facets of L M that are in the first layer connect through the nonempty base.The combination of these two facts implies that L M is connected, which will end our proof.
Claim 1: (The second layer facets of L M intersect the first layer facets).For any i ∈ [q] we have: Since M = m i r , Proposition 5.6 guarantees that there exists j ∈ [q] with j = i and m j | M so that (r − 1)e i + e j ∈ V .
Since m i r | M and m j | M , we see that m i r−1 m j | M .Indeed, we set M = m i r M ′ and, since m j is square-free, we have Claim 2: (The first layer facets connect through the base of L M ).For any i, j ∈ [q] with i = j we have: Proof of Claim 2. Assume for some i, j ∈ [q] with i = j that Assume, by way of contradiction, that Since b j > s ≥ r − s, by Lemma 5.7 (ii) and Remark 5.8, Moreover c ∈ B r , because If L = L r q , then V = N r q .Since (5.9.3) shows c ∈ V M , we conclude c ∈ B r ∩ V M , a contradiction.This finishes the proof of part (1).
Assume now L = L r (I).Since we assumed contradicting the assumption made in (5.9.2).This finishes the proof of part (2).
Remark 5.10.One may feel that part (1) of the statement above is weaker than (2), since L r (I) ⊆ L r q .However, the remarkable aspect of (1) is that, before labeling, L r q does not depend on the ideal I. Thus the same topological structure, depending only on r and q, supports a free resolution of the r th power of any ideal generated by q square-free monomials.The idea is that L r q provides an alternative notion to the Taylor complex for powers of I: when r = 1, L r q is the Taylor simplex, and when r > 1, L r q is significantly smaller than the Taylor simplex but still supports a resolution of I r .

BOUNDS ON BETTI NUMBERS
One of the key applications of Theorem 5.9 is that we are able to improve the bounds on Betti numbers for powers of ideals from that given by the standard Taylor resolution of I r since L r (I) ⊆ L r q ⊆ Taylor(I r ).This section contains computations that illustrate the extent to which our results improve the Taylor bounds.
When r = 2, we were able to provide a concrete formula for the number of faces of L r (I) in [4], and as a result we provided bounds for Betti numbers of I 2 , which are shown in Proposition 7.11 to be sharp.When r > 2, however, such formulas are not as easy to write and the numbers are very large even for small examples.In this case we shift our attention to L r q .While the bound on the Betti numbers stemming from L r (I) is dependent on the relations among the generators of I, one can use the structure of L r q to get a general, albeit weaker, bound on β t (I r ).Theorem 6.1 (Bounds on Betti numbers of I r ).If I is a square-free monomial ideal with q minimal generators, then the Betti numbers of I r for r ≥ 2 are bounded above by where t ≥ 0, b is the coefficient of x r in the expansion of In particular, pd(I r ) ≤ max{q − 1, f − 1}.
Proof.Let s = r 2 .Note that β t (I r ) is bounded above by the number of faces of dimension t of L r q .To count the faces of dimension t, we use the sets B r , F r i , and G r i from Definition 4.2.Observe that the coefficient b of x r in the expansion of is exactly the number of q-tuples (a 1 , . . ., a q ) with a i ≤ s and Note that the vertices of L r q are formed by selecting r of the original q generators, so using the standard formula for counting with repetition, we have , there are |V (L r q )| − q vertices in ∪ q i=1 F r i .Also F r i ∩ F r j = B r for all i, j such that i = j, so we have To count the number of faces of dimension t, that is, faces with t + 1 vertices, in L r q , we separate the faces into three distinct types.
(1) Faces containing m i r for some i: By the definition of L r q , faces of this type must be contained in G r i .Since the vertex corresponding to m i r has been fixed, t additional vertices of G r i are needed.There are q − 1 such vertices since all vertices of G r i have the form (r − 1)e i + e j where 1 ≤ j ≤ q.Since there are q choices for i, there are q q − 1 t such faces.(3) Faces contained in F r i but not in B r : Recalling that f = |F r i | and B r ⊆ F r i , there are Combining the three cases, we see that there are q of dimension t.Thus for I r , In particular, if t > q − 1 and t + 1 > f , we must have β t (I r ) = 0. Thus Corollary 6.2.If an ideal I is minimally generated by q square-free monomials, then the Betti numbers of I r are bounded by , where b is as defined in Theorem 6.1.In the case where r = 2, b = q 2 and the bound reduces to the bound given in [4].
Proof.When r = 2 or r = 3, then F r i = B r for all i, so b = f and the result follows immediately from simplifying the formula in Theorem 6.1.Moreover, when r = 2, the coefficient of x 2 in the binomial expansion of (1 + x) q is q 2 as stated.The reduction is then evident.Notice that when r = 2, the bound is known to be sharp; it agrees with the result in [4].In Proposition 7.11 we characterize the values of r and q for which these bounds are sharp, i.e. can be realized by some ideal.Example 6.3.As a first example, we examine cases where r or q is small.We first consider b and f for small values of q.
• For q = 2, computing (1 shows that b = 1 when r = 2s is even and b = 2 when r = 2s − 1 is odd.If r is even, we then have A similar computation shows that when r is odd, f = s as well.• If q = 3 and r = 3 or r = 4 (so that s = 2), then computing (1 + x + x 2 ) 3 yields b = 7 and f = 7 when r = 3 and b = 6 and f = 8 when r = 4. Applying the equations from Theorem 6.1 yields: • For r = 2, any q, and any t, • For any r, q = 2, and any t ≥ 2, if s = ⌈r/2⌉, β t (I r ) ≤ 2 s t + 1 .
• For r = q = 3 and any t, • For r = 4, q = 3 and any t ≥ 3, Remark 6.4.In (6.1.2) it is useful to understand which of the integers q − 1, f − 1 achieves the maximum.
For small values of q and r, it is possible to have q ≥ f .However, the opposite holds in most cases.More precisely, we show below that if r, q satisfy any of the following assumptions: (1) q = 2 and r ≥ 5; (2) q = 3 and r ≥ 3; (3) q ≥ 4 and r ≥ 2, then f > q, and hence pd(I r ) ≤ f − 1 = dim L r q .In the case q = 2, (6.3.1)shows f = s = ⌈ r 2 ⌉.If r ≥ 5, we must have s > 2, thus f = s > 2 = q.This settles (1).Now suppose q > 2. If r = 2, then Corollary 6.2 shows that f = b = q 2 .When q ≥ 4, we have f = q 2 > q.Thus, to show (2) and ( 3) it remains to consider the case when q ≥ 3 and r ≥ 3. Observe that when r ≥ 3, s ≥ 2. In this scenario, since b is the coefficient of x r in (1 + x + • • • + x s ) q and q > 2, then b ≥ 2 and bq − b > q.
To see that f > q holds, we will show f − q > 0, which by (6.1.1)amounts to It is sufficient to show the numerator is positive, so we observe the following, where the first inequality results from bq − b > q and the second follows since r ≥ 3: = (q + 2)(q + 1)q 6 − q 2 = q(q − 1)(q − 2) 6 > 0 .
This ends our argument.
Example 6.5.In general, the bounds from Theorem 6.1 will be considerably smaller than those provided by the Taylor complex, which is a simplex, and the bound on the projective dimension will also decrease significantly.We display this phenomenon in the table below.
Bound Comparisons q = 2, r = 3 q = 3, r = 3 q = 3, r = 4 Theorem 6.The bounds on Betti numbers and projective dimension given by the complex L r q in Theorem 6.1 are most effective when the generating set {m a | a ∈ N r q } does not contain redundancies.When there are redundancies, using L r (I) will yield improved bounds.We illustrate how to use this improvement by continuing Example 5.2.Example 6.6.Let I = (xy, yz, zw, xw) = (m 1 , m 2 , m 3 , m 4 ) as in Example 5.2.Counting faces of size i in the complex L 2 (I) provides a bound on the i th Betti number of I 2 .Note that in general, these improvements in the Betti numbers follow from knowledge of the redundancies in {m a | a ∈ N r q } and can often be computed from that information using the equivalence classes without explicitly constructing L r (I).For instance, a comparison of the bounds is summarized in the table below for I 2 and for I 3 using the first of the two vertex sets for L 3 (I) given in Example 5.2.Based upon the above work, a natural question that arises is the following: for given r and q, can one find ideals I for which I r has a minimal free resolution supported on L r q itself?When r = 1, L r q is the Taylor complex, which one can easily see always supports a minimal free resolution of the ideal generated by q variables (x 1 , x 2 , . . ., x q ).In the case where r > 1, Proposition 7.11 describes exactly when the bounds for Betti numbers in Theorem 6.1 and Corollary 6.2 are sharp, in the sense that there exist ideals for which equality is attained.We call the ideals which realize these bounds q-extremal ideals and denote them by E q .In fact we can show a much stronger statement in Theorem 7.9: the powers E q r have maximal Betti numbers among the ideals I r where I is generated by q square-free monomials.Definition 7.1 (Extremal ideals).Let q be a positive integer.For every set A with ∅ = A [q], introduce a variable x A , and consider the polynomial ring

Bound Comparisons
over a field k.For each i ∈ [q] define a square-free monomial ǫ i in S E as The square-free monomial ideal E q = (ǫ 1 , . . ., ǫ q ) is called a q q q-extremal ideal.
When it is unlikely to lead to confusion, we will simplify the notation by writing x 1 for x {1} , x 12 for x {1,2} , etc., and refer to a q-extremal ideal simply as an extremal ideal.The ring S E has 2 q − 1 variables, corresponding to the power set of [q] (excluding ∅), and each ǫ i is the product of 2 q−1 variables; that is, those corresponding to the subsets of [q] that contain i.The following example illustrates how this works for E 4 .
We show in Theorem 7.9 that powers of extremal ideals attain maximal Betti numbers among powers of all square-free monomial ideals with the same number of generators.To prove this, we first define a ring homomorphism S E → S and discuss its properties.Before proceeding directly with our work on extremal ideals, we illustrate how the map ψ I works in a sample case where there are four generators and seven variables.Generalizing the properties of ψ I from the example, we arrive at the following lemma.
Set a i = (a i1 , a i2 , . . ., a iq ), for i ∈ [t].Using (7.4.1) in the first equality below, we have: m atj j = lcm(m a1 , . . ., m at ). Theorem 7.9 below demonstrates why the ideals from Definition 7.5 are called extremal: they have the greatest Betti numbers among all ideals minimally generated by q square-free monomials.The following lemma provides the technical preliminaries necessary for the proof of Theorem 7.9.
Lemma 7.8.Let I be an ideal minimally generated by q square-free monomials in a polynomial ring k[x 1 , . . ., x n ].With notation as in Definition 7.5, if S is viewed as an S E -module via the ring homomorphism ψ I : S E → S, then S E /E q r ⊗ SE S ∼ = S/I r and Tor SE i (S E /E q r , S) = 0 for all i > 0.
Proof.First, note that S E E q r ⊗ SE S ∼ = S (E q r )S = S ψ I (E q r )S = S I r .
To compute Tor SE i (S E /E q r , S), use the Taylor complex Taylor(E q r ), which supports a free resolution F of S E /E q r ; see (2.4.1) for a description of the differentials of F. Since the homomorphism ψ I changes the labels ǫ ǫ ǫ a to the labels m a and preserves least common multiples, the chain complex F ⊗ SE S is isomorphic to a homogenization of the chain complex associated to the simplex with vertices corresponding to a ∈ N r q and labeled with the monomials m a .This is a version of the Taylor resolution of S/I r defined starting with a possibly non-minimal generating set of I r .Such a non-minimal version is a free resolution of S/I r as well, hence Tor SE i (S E /E q r , S) = 0 when i > 0. (See Remark 2.6.)Theorem 7.9 (Powers of extremal ideals have maximal Betti numbers).Given positive integers r and q, β S i (I r ) ≤ β SE i (E q r ) for any i ≥ 0 and any ideal I minimally generated by q square-free monomials in a polynomial ring S. Now assume q ≥ 3 and 1 ≤ r ≤ 2. We will show that, for every c ∈ N r q and σ ∈ L r q , ǫ ǫ ǫ c | M σ ⇐⇒ c ∈ σ. (7.10.5) When r = 1 observe that for every i ∈ [q], ǫ i ∤ lcm(ǫ k1 , . . ., ǫ kp ) for all i ∈ [q], k 1 , . . ., k p ∈ [q] {i}.(7.10.6)This can be seen by noting that the right-hand side of (7.4.2) becomes 1 ≤ 0 for A = {i}, ǫ ǫ ǫ c = ǫ i and σ = {ǫ k1 , . . ., ǫ kp }.Therefore L 1 q = L 1 (E q ) = Taylor(E q ), and (7.10.5)follows immediately.

Example 5 . 2 .
Let S = k[x, y, z, w] and I = (xy, yz, zw, xw) = (m 1 , m 2 , m 3 , m 4 ).By Proposition 4.3, the facets of L 2 4 are a 5-simplex B 2 and four tetrahedra G 2 i for 1 ≤ i ≤ 4, depicted on the left in the figure below.By Definition 5.1, since m 1 m 3 = m 2 m 4 is the only equation determining an equivalence class with more than one element, we select the vertex e 1 + e 3 to represent this equivalence class and form L 2 (I).Then L 2 (I) consists of a 4-simplex on the vertices e 1 + e 2 , e 1 + e 3 , e 1 + e 4 , e 2 + e 3 , e 3 + e 4 together with two triangles and two tetrahedra depicted on the right in the figure below.Note that vertex e 2 + e 4 has been removed.The edges depicted by dotted lines in the figure of L 2 4 do not appear in L 2 (I) and higher dimensional faces of L 24 containing e 2 + e 4 have also been deleted.
re i and (r − 1)e i + e j are vertices of L r (I), and m i r and m i r−1 m j are minimal monomial generators of I r .
and we conclude u = v because m u , m v are minimal generators.
.6.6), together with the fact that m c and m d are square-free, imply that m c | M and m d | M, which in turn implies that c, d ∈ A. Assume without loss of generality that deg m ′ c ≤ deg m ′ d .

( 2 )
Faces contained in B r : Setting b = |B r | as above, there are b t + 1 such faces.

Definition 7 . 5 (
The ring homomorphism ψ I ).Let I be an ideal of the polynomial ring S = k[x 1 , . . ., x n ] minimally generated by square-free monomials m 1 , . . ., m q .For each k ∈ [n], setA k = {j ∈ [q] : x k | m j } .We have thus j ∈ A k ⇐⇒ x k | m j .Define ψ I to be the ring homomorphismψ I : S E → S where ψ I (x A ) = k if A = A k for some k ∈ [n],1otherwise.