On Monomial Golod Ideals

We study ideal-theoretic conditions for a monomial ideal to be Golod. For ideals in a polynomial ring in three variables, our criteria give a complete characterization. Over such rings, we show that the product of two monomial ideals is Golod.


Introduction
Let k be a field, and let Q = k[x 1 , . . . , x n ] be a polynomial ring on n variables over k, with deg(x i ) = 1 for all i. We denote by m = (x 1 , . . . , x n ) the homogeneous maximal ideal of Q. Let I ⊆ m 2 be a homogeneous ideal and R = Q/I . Serre proved a coefficient-wise inequality of formal power series for the Poincare series of R: When equality happens, the ring R (and the ideal I ) is called Golod. The notion is defined and studied extensively in the local setting, but in this paper we shall restrict ourselves to the graded situation. Golod rings and ideals have attracted increasing attention recently (see [6,8,9,12,15]), but they remain mysterious even when n = 3. For instance, we do not know if the product of any two homogeneous ideals in Q = k[x, y, z] is Golod. Another reason for the increasing interest is their connection to moment-angle complexes (for example, see [7,10,14]).
It was asked by Welker whether it is always the case that the product of two proper homogeneous ideals is Golod (for example, see [16,Problem 6.18]) but a counter-example, even for monomial ideals, was constructed by the second author in [8].
In this work, we provide a concrete characterization of Golod monomial ideals in three variables, and use it to show that the product of any two proper monomial ideals in Q = k[x, y, z] is Golod. The following is our first main result: (2) [I : We point out that rings with embedding codepth at most three have been studied extensively. For instance, their Koszul homology has been completely classified up to isomorphism (see [1,[3][4][5]18] Our second main result is a consequence of Theorem 1.1: we obtain that products of monomial ideals in three variables are Golod.
Observe that this result is optimal, as the example of a non-Golod product of two monomial ideals constructed in [8] is in four variables. We end the paper with some positive and negative partial results regarding the colon conditions highlighted by this work, and several open questions.

Characterization of Monomial Golod Ideals in Three Variables
In this section, we prove Theorem 1.1. We first focus on the necessary part, which holds quite generally. Let Q = k[x 1 , . . . , x n ], I a homogeneous ideal in Q, and R = Q/I . Let K Q be the Koszul complex on a minimal set of generators x 1 , . . . , x n of the maximal ideal m of Q, and K R = R ⊗ Q K Q . The Koszul complex can be realized as the exterior algebra K R 1 , where K R 1 is a free R-module of rank n, with basis e x 1 , . . . , e x n . An element of the p-th graded component K R p can be written as a sum of elements of the form re x i 1 ...x ip , where 1 ≤ i 1 < i 1 < · · · < i p ≤ n, r ∈ R, and where we set e x i 1 ...x ip := e x i 1 ∧ . . . ∧ e x ip . The Koszul complex also comes equipped with a differential ∂, as it is a DG algebra. The differential is such that ∂(e x i 1 ...  Similarly, take f ∈ I : (x 1 , . . . , x p ) and g ∈ I : (x p+1 , . . . , x n−1 ). Consider the cycles f e x 1 ...x p and ge x p+1 ...x n−1 . The product of these is zero in H n−1 (R) if and only if there is h ∈ R such that ∂(he x 1 ...x n ) = fge x 1 ...x n−1 . But this means that hx i = 0 for 1 ≤ i < n and hx n = fg in R. Lifting to Q, this shows that h ∈ I : (x 1 , . . . , x n−1 ), and thus fg ∈ x n [I : [8]. It can be used to easily provide examples of non-Golod ideals. For example, let

Remark 2.2 The above proposition is motivated by the examples in
Then, xy ∈ I : (x, y) and zt ∈ I : (z, t) but xyzt / ∈ I . Thus, I is not Golod.
It is well-known that, for homogeneous ideals inside polynomial rings in three variables, being Golod is equivalent to requiring that the product on the Koszul homology is trivial (for instance, see [15,Theorem 6.3]). In the same article, it is shown that this is not the case more generally, even for monomial ideals. In order to prove the converse of Proposition 2.1 for monomial ideals in k[x, y, z], we show that the Koszul homology modules admit "monomial bases". This is what we shall focus on for the rest of this section.

Definition 2.3 Let
. , x n ], and I ⊆ m 2 be a monomial ideal. Let R = Q/I . We say that H p (K R ) admits a monomial basis if it has a k-basis consisting of classes of cycles of the form ue x i 1 ...x ip , where u ∈ Q is a monomial and u denotes its image inside R.
Observe that, if the ideal I is homogeneous, then we can talk about homogeneous ele- In this case, the differential preserves degrees. Even more specifically, if I is monomial, then each K R p is a Z n -graded R-module. If r = x a 1 1 · · · x a n n , then re x i 1 ...x ip has degree (a 1 , . . . , a n ) + i 1 + · · · + i p , where j is the vector in Z n which has 1 in position j and 0 elsewhere. For example, x 2 y 3 e yz ∈ K k[x,y,z] 2 has degree (2, 4, 1). In this case, the differential ∂ on K R preserves multidegrees.
The following is well-known. Nonetheless, we provide a short proof for completeness. Proof Since I is a monomial ideal, R admits a graded free resolution with Z n -graded shifts (for example, the Taylor resolution). There is a Z n -graded isomorphism H p (K R ) ∼ = Tor Q p (Q/I, k) that comes from tracing Koszul cycles along the double complex P • ⊗ K Q , where P • → R is a Z n -graded free resolution of R, and K Q can be viewed as a Z n -graded minimal free resolution of Q/m ∼ = k. Since Q/I has Z n -graded shifts, we see that Tor Q p (Q/I, k) has a Z n -graded k-basis. Via this isomorphism, such a basis maps to a set of graded Koszul cycles in K R p , which forms a k-basis in homology.
We observe that if Proof It is clear that a k-basis of H n (K R ) can be chosen to be of such form. In fact, an element of K R n is of the form f e x 1 ...x n , where f ∈ I : m. Since I is monomial, we can choose f to be a monomial.
It is also fairly easy to prove the claim for H 1 (K R ) directly. However, we explain the process via lifting Koszul cycles, as this technique will be used later. Let (P • , δ) be a graded free resolution of R as a Q-module. As noted in Proposition 2.4, we have a graded isomorphism between Tor Proof The statement for H 1 (K R ) and H 3 (K R ) has already been proved in Lemma 2.5 (assuming that depth(R) = 0 for the latter to be non-zero). The argument for H 2 (K R ) exploits again the process of lifting Koszul cycles. Assume that H 2 (K R ) = 0, that is, depth(R) ≤ 1. We consider a minimal Z 3 -graded free resolution of R over Q After fixing bases, δ 1 can be represented as the matrix [m 1 , . . . , m t ], where {m 1 , . . . , m t } is a minimal monomial generating set of I . On the other hand, δ 2 is represented by a matrix where every column has precisely two non-zero monomial entries. This is because every relation between distinct monomials m i and m j is of this form m i u i − m j u j = 0 for some monomials u i , u j ∈ m. We now describe the lifting process for H 2 (K R ). Consider the following part of double complex F • ⊗ K Q • : If we lift a Z 3 -graded basis element of k(−b ) ⊆ Tor R 2 (Q/I, k) to ⊕ Q(−b ) ⊗ Q, this will map down via δ 2 ⊗ 1 to an element (0, . . . , u i , . . . , −u j , . . . , 0) ∈ ⊕ j Q(−a j ) ∼ = j Q(−a j ) ⊗ Q, corresponding to a binomial relation in the -th column of the matrix representing δ 2 , as described above. Write u i = x i v i and u j = x j v j , for some i, j , and monomials v i , v j ∈ Q. Observe that i = j , since otherwise the relation between m i and m j given by (0, . . . , u i , . . . , −u j , . . . , 0) would not be minimal. We may assume that i < j. Using the above relations, we have that (0, . . . , u i , . . . , −u j , . . . , 0) = (1 ⊗ ∂ 1 )(0, . . . , v i e x i , . . . , −v j e x j , . . . , 0). We now push this element down via δ 1 ⊗ 1, to get an we claim that (1 ⊗ ∂ 2 )(w ⊗ e x i x j ) = σ . By the definition of the differential, we have 2 , the process of lifting Koszul cycles now ends by considering the class of the element we x i x j inside H 2 (K R ). As observed above, inside Q we have wx i = −v j m j ∈ I , and wx j = −v i m i ∈ I . Therefore w ∈ I : (x i , x j ), and the class of we x i x j inside H 2 (K R ) gives then a basis element of the desired form.  (2) in the classification of the Koszul homology of rings with embedding codepth at most three. In particular, R is not Golod, by [3, 1.4.3]. However, it can be checked using Macaulay 2 [11] that the ideal g 1 satisfies conditions (1) and (2) of Theorem 1.1.
We observe that the condition that I ⊆ m 2 in Theorem 1.1 cannot be removed. We conclude this section observing that, in the case of monomial ideals in four or more variables, some Koszul homology modules may not always admit a "monomial basis." Example 2.10 Consider the ideal I = (xz, xw, yz, yw) in Q = k[x, y, z, w], and let R = Q/I . It is easy to check that α = xe yzw − ye xzw is a cycle of K R 3 , whose class equals that of we xyz − ze xyw in homology. Observe that α has multidegree (1, 1, 1, 1). Exploiting the multigrading, one can show that the class of α in homology cannot be expressed as a combination of elements coming from a monomial basis of H 3 (K R ).

Products of Monomial Ideals in k[x, y, z] are Golod
In this section, we prove Corollary 1.2. By Theorem 1.1 and by the symmetry, it suffices to show the following lemmas. Proof Let f ∈ I : x and g ∈ I : (y, z) be monomials. If f ∈ (y, z), then fg ∈ I , so we assume f = x a for some a ≥ 0. It follows that x a+1 ∈ J K, and as J, K are proper we must have f ∈ J ∩ K. As J, K are monomial ideals, we have g ∈ J K : y ⊆ (J : y)K + (K : y)J ⊆ J + K, and we are done.

Integrally Closed Ideals and Some Questions
The colon conditions considered in this paper seem related to the property of "being integrally closed" (see also the m-full and basically full conditions [13,17]). Here, we give some positive and negative results in this direction.  (x b , y, z). Because of our assumptions, we have a ≥ 1 and b ≥ 2. Let f ∈ J K : x, so that f x ∈ J K ⊆ (J, y, z)(K, y, z) ⊆ (x a+b , y, z).
It follows that f ∈ (x a+b−1 , y, z), write f = ux a+b−1 + v for some v ∈ (y, z). Now let g ∈ J K : (y, z), and consider the element h = ux b−1 g. As we are assuming b ≥ 2, we have h ∈ mJ K : (y, z). On the other hand, since we have f x = ux a+b + xv ∈ J K, we get that x a+1 h = ux a+b g = f xg − xvg ∈ mJ K. It follows that h ∈ mJ K : (x a+1 , y, z). Observe that (mJ, y, z) = (x a+1 , y, z). It follows that h ∈ mJ K : (x a+1 , y, z) ⊆ mJ K : mJ = K, because K is integrally closed. Since f is congruent to ux a+b−1 modulo (y, z), we have that fg is congruent to ux a+b−1 g = x a h modulo J K. On the other hand, x a belongs to (J, y, z); therefore, x a h ∈ (J, y, z)[K ∩ (J K : (y, z))] ⊆ J K. Unfortunately, one can use other necessary colon criteria provided in Proposition 2.1 to show that even integrally closed monomial ideal in three variables or product of them in four variables may not be Golod.  We do not know the answer to Question 4.5 even when K = m. One can show that the conclusions of Lemmas 3.1 and 3.2 still hold when K = m and J is any proper homogeneous ideal in Q, so Proposition 2.1 does not provide any obstructions in this case. When the characteristic of k is 0 and J = K, the answer is positive by the main result of [12].
Finally, we have not been able to determine whether Lemma 3.1 holds for any product J K of homogeneous ideals in three variables, without either the monomial condition, or assuming that K ⊆ m 2 is integrally closed. Observe that any example for which the lemma fails would provide a negative answer to Question 4.5. It is rather frustrating that such a simple-looking question cannot be resolved, so the first author is willing to offer a cash prize of 25 USD for the first solver of this: Funding Open access funding provided by Università degli Studi di Genova within the CRUI-CARE Agreement. The first author is partially supported by NSA grant H98230-16-1-001 during the preparation of this work.
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