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enrico.sbarra@unipi.it Dipartimento di Matematica, In this survey paper, we ﬁrst present the main properties of sequentially Cohen–Maca Università degli Studi di Pisa, ulay modules. Some basic examples are provided to help the reader with quickly gett Largo Bruno Pontecorvo 5, 56127 Pisa, Italy ing acquainted with this topic. We then discuss two generalizations of the notion of Full list of author information is sequential Cohen–Macaulayness which are inspired by a theorem of Jürgen Herzog available at the end of the article and the third author. Keywords: Sequentially Cohen-Macaulay module, Dimension ﬁltration, Generic initial ideals, Local cohomology Introduction This survey note, which is dedicated to the work of Jürgen Herzog on the topic, cannot possibly be complete: the notion of sequentially Cohen–Macaulay module has been central in many papers in the literature from the late 90’s, starting maybe with [20]. On the other hand, in the late 90’s Herzog’s research activity was feverish as he counted very many visitors and collaborators. At the same time, the distribution of preprints in the form of postscript ﬁles over the internet expedited the dissemination of mathematical ideas. We thus must apologize in advance that our reference list is doomed to be incomplete. It is in 1997 that Herzog, together with who would become his top co-author, Takayuki Hibi, published a paper on simplicial complexes [26], immediately followed by another series of papers of the two authors together with Annetta Aramova [5–7], where numer- ical problems about simplicial complexes were addressed through the study of Hilbert functions, Gröbner bases techniques and generic initial ideals, see also [3,4,19]. In 1999, another inﬂuential paper authored by Herzog and Hibi, is published, [27]: they intro- duce and study a new class of ideals, called componentwise linear. Componentwise linear Stanley–Reisner ideals I are characterized as those for which the pure i-th skeleton of the Alexander dual of is Cohen–Macaulay for every i.Thanksto[20], this means that I is componentwise linear if and only its Alexander dual is sequentially Cohen–Macaulay, see also [30]. In this way, the authors were also able to generalize a well-known result of Eagon and Reiner, which says that the Stanley–Reisner ideal of a simplicial complex has a linear resolution if and only if its Alexander dual is Cohen–Macaulay. All the ideas behind the proofs of these facts led also to another fundamental result, which is the main theo- rem of [31] and provides a somewhat unexpected characterization of graded sequentially © The Author(s) 2022. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. 0123456789().,–: volV 40 Page 2 of 27 Caviglia et al. Res Math Sci (2022) 9:40 Cohen–Macaulay modules over a polynomial ring R in terms of the Hilbert function of the local cohomology modules: Theorem 1 Let M be a ﬁnitely generated graded R-module, and let M = F /Ua free graded presentation of M. Then, F /U is sequentially Cohen–Macaulay if and only if i i Hilb(H (F /U)) = Hilb(H (F / Gin(U))) for all i 0. m m Here, Gin(U) denotes the generic initial module of U with respect to the degree reverse lexicographic order. This paper is organized as follows. In the ﬁrst section, we introduce the deﬁnition of sequentially Cohen–Macaulay modules according to Stanley [37], and discuss some basic properties and examples. In Sect. 2, we present some of the main characterizations, or equivalent deﬁnitions, of sequential Cohen–Macaulayness, by recalling Schenzel’s results on the dimension ﬁltration of a module, cf. Propositions 3, 5 and Theorem 2, and Peskine’s characterization in terms of deﬁciency modules, cf. Theorem 3. In Sect. 3, we recollect the deﬁnition of partial sequential Cohen–Macaulayness introduced by the third and the fourth author in [35], by requiring that only the queue of the dimension ﬁltration is a Cohen–Macaulay ﬁltration, see Deﬁnition 5 and some basic properties in the graded case in Lemmata 2 and 3. The next two results, Lemma 4 and Proposition 8, are dedicated to ﬁlling a gap in the proof of a fundamental lemma in [35], and we present the ﬁrst of our generalizations of Theorem 1 in Theorem 4. We start Sect. 4 by recalling the deﬁnition of E-depth, as introduced by the ﬁrst and second author in [12], see Deﬁnition 7.E-depth measures how much depth the deﬁciency modules of a ﬁnitely generated standard graded module M have altogether, and sequential Cohen–Macaulayness can be detected by E- depth as observed in Remark 10. Some interesting homological properties of E-depth, especially in connection with strictly ﬁlter regular elements, are shown in Proposition 9. In Deﬁnition 9, we introduce the other crucial notion for the following, what we might call partial generic initial ideal, making use of a special partial revlex order. Finally, by means of Proposition 11, we prove in Theorem 7 the main result of this section, which can be seen as another generalization of Theorem 1. To Jürgen Herzog, a bright example of mathematician and teacher. 1 Sequentially Cohen–Macaulay modules Throughout the paper, let (R, m,k) denote either a Noetherian local ring with maximal ideal m and residue ﬁeld k = R/m, or a standard graded k-algebra R = R ,with i0 R = k and m = R . In the second case, every R-module we consider will be a graded 0 i i>0 R-module, and homomorphisms will be graded of degree zero. One of the features that makes the Cohen–Macaulay property signiﬁcant is its char- acterization in terms of the vanishing and non-vanishing of local cohomology: for a d- dimensional ﬁnitely generated module M with t = depth M,itholds that H (M) = 0 for t d all i < t and i > d;also, H (M) = 0and H (M) = 0. These results are originally due m m to Grothendieck, cf. [10], Theorem 3.5.8, Corollary 3.5.9, Corollary 3.5.11.a) and b). As a consequence, M is a Cohen–Macaulay module if and only if H (M) = 0 for all i = d. m Caviglia et al. Res Math Sci (2022) 9:40 Page 3 of 27 40 Let dim(R) = n. When R is Cohen–Macaulay and has a canonical module ω ,by duality, the above conditions can be checked on the Matlis dual of the local cohomology n−i modules, i.e., on the modules Ext (M, ω ). With the above notation, we then have that n−t n−d Ext (M, ω )and Ext (M, ω ) are nonzero, and furthermore that Ext (M, ω ) = 0 R R R R R R for all i < n − d and all i > n − t. Moreover, M is Cohen–Macaulay if and only if n−i n−i Ext (M, ω ) = 0 for all i = d. Often in the literature, the modules Ext (M, ω ), R R R R i = 0, ... ,n, are called the deﬁciency modules of M, since they also measure how far a module is from being Cohen–Macaulay. We now introduce the class of sequentially Cohen–Macaulay modules, following [37], recalling their main properties, and in the next section, we discuss some of its equivalent deﬁnitions. Our main references here are [31]and [36]. Deﬁnition 1 A ﬁnitely generated R-module M is called sequentially Cohen–Macaulay if it admits a ﬁltration of submodules 0 = M M ... M = M such that each quo- 0 1 r tient of the ﬁltration M /M is Cohen–Macaulay and dim(M /M ) < dim(M /M ) i i−1 i i−1 i+1 i for all i. In this case, we say that the above ﬁltration is a sCM ﬁltration for M. Example 1 (1) It is clear from the deﬁnition that if M = 0 is a Cohen–Macaulay module, then it is sequentially Cohen–Macaulay. In fact, M = M M = 0. 1 0 (2) Any one-dimensional module M is sequentially Cohen–Macaulay; if M is not Cohen–Macaulay, one can take M = H (M)and M = M. 1 2 (3) A domain R is sequentially Cohen–Macaulay if and only if it is Cohen–Macaulay. In fact, any nonzero submodule of R will have dimension d = dim(R), and the only possible sCM ﬁltration for R is 0 = M M = R. 0 1 (4) On the other hand, if M has a Cohen–Macaulay submodule M such that M/M 1 1 is sequentially Cohen–Macaulay with a ﬁltration 0 = M /M N /M ... 1 1 1 1 N /M = M/M ,and dim M < dim N /M ,then0 M N ... N = M s 1 1 1 1 1 1 1 s is a sCM ﬁltration of M. The notion of sequentially Cohen–Macaulay modules appears for the ﬁrst time in the literature in [37] in the graded setting, in connection with the theory of Stanley–Reisner rings and simplicial complexes. Later on and independently, in [36], the notion of Cohen– Macaulay ﬁltered modules has been introduced in the local case; in the same paper, it is proven that the two notions coincide. Since then, sequentially Cohen–Macaulay modules have been extensively studied especially in connection with shellability and graph theory, see for instance [1,2,22,25,40], and the deﬁnition has been extended in other directions, see [14], [33]. Sections 3 and 4 are devoted to two such extensions, due to the third and fourth author and the ﬁrst two authors, respectively. Example 2 Let M be a sequentially Cohen–Macaulay module with sCM ﬁltration 0 = M M ··· M = M and let d = dim(M /M ) for all i. 0 1 r i i i−1 (1) Let N be another sequentially Cohen–Macaulay; then, M ⊕N is sequentially Cohen– Macaulay. To see this, let 0 = N N ... N = N be a sCM ﬁltration of 0 1 s N and let δ = dim(N /N ). For convenience, also let d = δ =−1. Then, a i i i−1 0 0 ﬁltration whose terms are of the form M ⊕ N , where either d = max{d | d δ } i j i a a j or δ = max{δ | δ d }, is a sCM ﬁltration of M ⊕ N = M ⊕ N. j a a i r s 40 Page 4 of 27 Caviglia et al. Res Math Sci (2022) 9:40 (2) The completion M of M at the maximal ideal m is a sequentially Cohen–Macaulay R-module. To this end, let M = M ⊗ R, and by ﬂatness of R,wehavethat0 = i R M ... M is a sCM ﬁltration of M ⊗ M R = M. 0 1 (3) It can be proven in general that being sequentially Cohen–Macaulay is preserved by faithfully ﬂat base changes; with some extra eﬀort, one can derive that M is a sequentially Cohen–Macaulay R-module if and only if M[|x|] is sequentially Cohen– Macaulay as an R[|x|]-module, cf [36, Theorem 6.2]. Remark 1 In the above example Part (1), the converse also holds, cf. [16, Proposition 4.5], [39, Proposition 3.2] or Corollary 4. In Part (2) of the same example, the vice versa does not hold in general, see for instance [36, Example 6.1]. It is true though when R is a homomorphic image of a Cohen–Macaulay ring, see [23, Corollaries 2.8] and [23, Corollaries 2.9], where it is proved that in such a case M is a sequentially Cohen–Macaulay R-module if and only if M is a sequentially Cohen–Macaulay R-module. Proposition 1 Let M be a sequentially Cohen–Macaulay R-module with sCM ﬁltration 0 = M M ... M = M. Also let d = dim(M /M ),for alli ∈{1, ... ,r},then: 0 1 r i i i−1 (1) for all j ∈ Z,wehavethat H (M) = 0 if and only if j ∈{d , ... ,d },and 1 r d d d i ∼ i ∼ i H (M) = H (M ) = H (M /M ); i i i−1 m m m (2) for all i ∈{1, ... ,r} and j < i, the modules M and M /M are sequentially Cohen– i i j Macaulay. Proof For Part (1), using the short exact sequences 0 → M → M → M /M → 0 i−1 i i i−1 one inductively shows that dim(M ) = d for all i ∈{1, ... ,r},since d < d .The i i i−1 i induced long exact sequences on local cohomology j−1 j j j ··· → H (M /M ) → H (M ) → H (M ) → H (M /M ) i i−1 i−1 i i i−1 m m m m j+1 → H (M ) → ··· i−1 together with the fact that M /M is Cohen–Macaulay of dimension d ,imply that i i−1 i j j d d i ∼ i ∼ H (M ) = H (M /M ),H (M ) = H (M )if j < d , i i i−1 m i m i−1 i m m and H (M ) = 0 otherwise. m i j j j In particular, H (M) = 0 for all j > d ,and H (M) = H (M ) for all j d . In conclusion, m r m m i i we have shown that d d ⎨ i i H (M ) = H (M /M )if j = d for some i, m i m i i−1 i H (M) = 0 otherwise. Part (2) is clear once we notice that 0 = M M ... M and 0 = M /M 0 1 i j j M /M ... M /M are sCM ﬁltrations. j+1 j i j Example 3 Suppose that x is an M-regular element and that M/xM is sequentially Cohen–Macaulay; it is not true in general that M is sequentially Cohen–Macaulay, as observed after [38, Proposition 2.2]—the statement of [36, Theorem 4.7] is not correct. As a counterexample, one can take a two-dimensional not Cohen–Macaulay local domain R of depth 1, which is not sequentially Cohen–Macaulay by Example 1 (3). For instance, take Caviglia et al. Res Math Sci (2022) 9:40 Page 5 of 27 40 4 3 3 4 2 2 2 2 3 2 R = Q[|a ,a b, ab ,b |] Q[|z ,z ,z ,z ]/(z z − z z ,z − z z ,z z − z z ,z − z z ). 1 2 3 4 2 3 1 4 2 1 4 3 3 4 3 2 2 1 On the contrary, every 0 = x ∈ R is regular, R/(x) is a one-dimensional R-module and, therefore, always sequentially Cohen–Macaulay by Example 1 (2). In the rest of this section, let S = k[x , ... ,x ], with the standard grading. Given a 1 n monomial ideal I ⊂ S,welet G(I) denote the monomial minimal system of generators of I and m(I) = max{i | x divides u for some u ∈ G(I)}. An important class of sequentially Cohen–Macaulay modules is given by rings deﬁned by weakly stable ideals. Deﬁnition 2 A monomial ideal I ⊆ S is said to be weakly stable if for all monomials u ∈ I and all integers i, j with 1 j < i n, there exists t ∈ N such that x u/x ∈ I, where is j i the largest integer such that x divides u. Observe that the condition of the previous deﬁnition is veriﬁed if and only if it is veriﬁed for all u ∈ G(I). In the literature, weakly stable ideals are also called ideals of Borel type, quasi-stable ideals, or ideals of nested type, see [8,11,28]. It is easy too see from the deﬁnition that stable, strongly stable, and p-Borel ideals are weakly stable. In particular, no matter what the characteristic of the ﬁeld is, Borel ﬁxed ideals are weakly stable, see also [28, Theorem 4.2.10] and, thus, generic initial ideals are always weakly stable. sat ∞ ∞ Observe that the saturation I of a weakly stable ideal I is I : m = I : x ,see [28, sat sat Proposition 4.2.9]; thus, I is again weakly stable and x does not divide any u ∈ G(I ). Proposition 2 Let I be a weakly stable ideal of S = k[x , ... ,x ].Then, S/I is sequentially 1 n Cohen–Macaulay. Proof We prove the statement by induction on m(I). If m(I) = 1, then I = (x ) for some positive integer r and S/I is Cohen–Macaulay. Assume now that S/J is sequentially Cohen–Macaulay for every weakly stable ideal J for which m(J) < m(I). Let S = k[x , ... ,x ]and I = I ∩ S .Since S/I S /I ⊗ k[x , ... ,x ]and a 1 m(I) k m(I)+1 n sCM ﬁltration of S /I is easily extended into one of S/I, it is suﬃcient to prove that S /I is sequentially Cohen–Macaulay. Therefore, without loss of generality, we may assume sat sat that m(I) = n, and that S/I is sequentially Cohen–Macaulay. Now, I /I is a non- sat trivial Artinian module, and the ﬁrst nonzero module of a sCM ﬁltration of S/I has positive dimension, see Proposition 1. We may thus conclude that S/I is sequentially Cohen–Macaulay, cf. Example 1 (4). Remark 2 By [8, Proposition 3.2], see also [28, Proposition 4.2.9] and [11,Chap. 4],a monomial ideal I is weakly stable if and only if all its associated primes are generated by initial segments of variables, i.e., are of type (x , ... ,x ) for some i. Since being sequentially 1 i Cohen–Macaulay is independent of coordinates changes on S, the above proposition shows that whenever the associated primes of a monomial ideal I ⊆ S are totally ordered by inclusion, then S/I is sequentially Cohen–Macaulay. 4 2 2 3 2 2 Example 4 Let I = (x ,x x ,x x ,x x x ,x x x ) ⊆ S = k[x ,x ,x ,x ]. It is easy to 3 2 3 2 4 1 2 3 4 1 1 2 1 1 1 verify that I is a weakly stable ideal which is not strongly stable, by checking the deﬁnition or by computing its associated primes, by using the previous remark; thus, S/I is sequen- tially Cohen–Macaulay by Proposition 2. We can construct an explicit sCM ﬁltration sat ∞ 4 2 3 proceeding as in its proof. We have m(I) = 4, and I = I : x = (x ,x x ,x x ). 2 3 4 1 1 1 40 Page 6 of 27 Caviglia et al. Res Math Sci (2022) 9:40 4 2 3 This means that the ﬁrst nonzero module of our ﬁltration will be (x ,x x ,x x )/I. 2 3 1 1 1 4 2 3 Now, we consider I = (x ,x x ,x x )asanideal of k[x ,x ,x ] and compute its sat- 2 3 1 2 3 1 1 1 ∞ 3 2 uration I = I : x = (x ,x x ). Hence, the second nonzero module of the ﬁltration 3 1 1 3 2 4 2 3 is (x ,x x )/I. Proceeding in this way, we obtain the ﬁltration 0 ⊆ (x ,x x ,x x )/I ⊆ 2 2 3 1 1 1 1 1 3 2 2 (x ,x x )/I ⊆ (x )/I ⊆ (1)/I = S/I, and it is easily seen that it is a sCM ﬁltration of S/I. 1 1 1 Example 5 For the case M = R = S/I, when I = I is the Stanley–Reisner ideal of a sim- plicial complex , there is a beautiful characterization of sequential Cohen–Macaulayness due to Duval, [20, Theorem 3.3]. Given a simplicial complex ,let (i) be the pure i-th skeleton of , i.e., the pure subcomplex of whose facets are the faces of of dimension i.Then, S/I is sequentially Cohen–Macaulay if and only if S/I is Cohen–Macaulay for (i) all i. Another important result in this context is that I is componentwise linear if and only if the Stanley–Reisner ring of its Alexander dual is sequentially Cohen–Macaulay, see [27, Theorem 2.1] and [28, Theorem 8.2.20]. Moreover, it is known that if is (nonpure) shellable, then S/I is sequentially Cohen–Macaulay, see [28, Corollary 8.2.19]. Example 6 Another class of examples of sequentially Cohen–Macaulay modules is given by pretty clean modules, which have been introduced by Herzog and Popescu in [29] in order to characterize shellability of multicomplexes. A pretty clean module M is a modulethatadmitsa pretty clean ﬁltration, i.e., a prime ﬁltration of M by submodules 0 = M M ... M = M such that each quotient M /M is isomorphic to S/p , 0 1 s i i−1 i for some prime ideals p , with the following property: if p p , then i > j. For example, i i j if M = S/I with I weakly stable, cf. [29, Proposition 5.2], or is such that Ass(M)isa totally ordered set, cf. [29, Proposition 5.1] or Remark 2, then M is pretty clean. By [29, Theorem 4.1], pretty clean modules are sequentially Cohen–Macaulay, provided every prime p appearing in the pretty clean ﬁltration is such that S/p is Cohen–Macaulay. i i 2 Characterizations of sequentially Cohen–Macaulay modules The goal of this section is to present two characterizations of sequentially Cohen– Macaulay modules, due to Schenzel, see Theorem 2 and Peskine, see Theorem 3.As in the previous section, we let (R, m,k) be either a Noetherian local ring or a standard graded k-algebra with maximal homogeneous ideal m;welet dim(R) = n. In the second case, modules will be graded and homomorphisms homogeneous of degree 0. 2.1 Schenzel’s characterization Our main reference here is [36]. By convention, the dimension of the zero module is set to be −1. Let M be a ﬁnitely generated (graded) R-module of dimension d.Since R is Noetherian, for all i = 0, ... ,d, we can consider the largest (graded) submodule of M of dimension i, and denote it by δ (M). By maximality, in this way, one obtains a ﬁltration M :0 ⊆ δ (M) ⊆ δ (M) ⊆ ... ⊆ δ (M) = M, called the dimension ﬁltration of M. 0 1 d Evidently, such a ﬁltration is unique. Given a set X of prime ideals of R,wedenoteby X ={p ∈ X :dim R/p = i}. Similarly, we deﬁne X and X . i >i Remark 3 Observe that M has a nonzero submodule of dimension i if and only if Ass(M) =∅.Infact, if p ∈ Ass(M) , then R/p is a nonzero submodule of M of dimension i i i. Conversely, if N is a nonzero submodule of M of dimension i, then there must exist a Caviglia et al. Res Math Sci (2022) 9:40 Page 7 of 27 40 minimal prime p of N such that dim(R/p) = dim(N) = i.Since N ⊆ M,wemustalso have that p ∈ Ass(M), as desired. In particular, we have that δ (M) = 0ifand only if Ass(M) =∅. Notice that δ (M) = H (M); the other modules in M can be described similarly, with the help of the minimal primary decomposition of 0 as a submodule of M.Infact, let Ass(M) ={p , ... , p } and, for all i = 0, ... ,d, consider the set Ass(M) .Weset 1 m i ⎨ p if Ass(M) =∅, p∈Ass(M) a = R otherwise. Now, let us consider an irredundant primary decomposition 0 = N inside M, j=1 where each M/N is a p -primary module. j j Proposition 3 [36, Proposition 2.2] Let M be a ﬁnitely generated R-module of dimension d. With the above notation, for all i = 0, ... , d we have that δ (M) = H (M) = N , i j {j | p ∈Ass(M) } j >i where we let the intersection over the empty set be equal to M. Proof We start with the ﬁrst equality and assume that δ (M) = 0. By Remark 3,wehave 0 0 that Ass(M) =∅ andinthiscase a = R.Thus, H (M) = H (M) = 0 = δ (M). i i i i R Now, assume that δ (M) = 0, and observe that every associated prime of δ (M)has i i dimension i and, therefore, Ass(δ (M)) ⊆ Ass(M) . Thus, there is a power of a i i i which annihilates δ (M), and we get that δ (M) ⊆ H (M). On the other hand, there is a i i 0 0 power of a which annihilates H (M)and,thus, dim(H (M)) dim(R/a ) i. From the i i a a i i maximality of δ (M), it follows that H (M) ⊆ δ (M), as desired. i i ∞ ∞ For the second equality, observe that N : p = M and N : x = N for any x ∈ / p , j M j M j j with j ∈{1, ... ,m},since M/N is p -primary. Also notice that j j ⎛ ⎞ m m 0 ∞ ∞ ∞ ⎝ ⎠ H (M) = 0: a = N : a = (N : a ). M j M j M a i i i j=1 j=1 ∞ ∞ Now, if p ∈ Ass(M) , then M ⊇ N : a ⊇ N : p = M, forcing equality j i j M j M i j everywhere. Next, assume that p ∈ / Ass(M) .Then, p p for every p ∈ Ass(M) j i j i since, otherwise, we would have dim(R/p ) dim(R/p) i and, thus, p ∈ Ass(M) .In j j i ∞ ∞ this case, by choosing x ∈ a and x ∈ / p , we have that N ⊆ N : a ⊆ N : x = N , i j j j M j M j and equalities hold. Summing up, we conclude that H (M) = N = N . j j {j | p ∈ /Ass(M) } {j | p ∈Ass(M) } j i j >i Note that this is consistent with our convention that the intersection over the empty set is equal to M; in fact, for i = d, we have that Ass(M) = Ass(M)and Ass(M) =∅. d >d 0 0 This agrees with the fact that H (M) = H (M) = M. Proposition 4 [[36], Corollary 2.3] Let M be a ﬁnitely generated R-module of dimension d; then, for all i = 0, ... ,d, 40 Page 8 of 27 Caviglia et al. Res Math Sci (2022) 9:40 (1) Ass(δ (M)) = Ass(M) ; i i (2) Ass(M/δ (M)) = Ass(M) ; i >i (3) Ass(δ (M)/δ (M)) = Ass(M) . i i−1 i 0 0 Proof It is well-known that Ass(M) = Ass(H (M)) Ass(M/H (M)), for any ideal a of a a R. Also notice that Ass(H (M)) = Ass(M) ∩ V (a), for all a. Since δ (M) = H (M)byProposition 3, the ﬁrst equality descends from Remark 3 and the above observation. For (2), consider the short exact sequence 0 −→ δ (M) −→ M −→ M/δ (M) −→ 0. i i By Proposition 3, one has that Ass(M) = Ass(δ (M)) Ass(M/δ (M)), which is equal to i i Ass(M) Ass(M) by (1), and the second equality follows. i >i Finally, consider the short exact sequence 0 → δ (M) → δ (M) → δ (M)/δ (M) → i−1 i i i−1 0 and observe that Ass(δ (M)/δ (M)) ⊆ Ass(M/δ (M)). We also have Ass(δ (M)/ i i−1 i−1 i 0 0 δ (M)) ⊆ Ass(δ (M)) since δ (M) = H (M) = H (δ (M)). Thus, by (1) and (2), i−1 i i−1 i a a i−1 i−1 we have necessarily that Ass(δ (M)/δ (M)) ⊆ Ass(M) . On the other hand, by (1), if i i−1 i p ∈ Ass(M) , then p ∈ / Ass(M ), and therefore p ∈ Ass(δ (M)/δ (M)). i i−1 i i−1 As a corollary of Propositions 3 and 4, one can obtain another characterization of the dimension ﬁltration, cf. [29, Proposition 1.1]. Corollary 1 Let M be a d-dimensional R-module. A ﬁltration of M by submodules 0 ⊆ M ⊆ ... ⊆ M = M is the dimension ﬁltration of M if and only if Ass(M /M ) = 0 d i i−1 Ass(M) ,for alli. The following deﬁnition was introduced in [36, Deﬁnition 4.1]. Deﬁnition 3 Let M be a d-dimensional ﬁnitely generated R-module, and M :0 ⊆ δ (M) ⊆ ... ⊆ δ (M) = M be its dimension ﬁltration. Then, M is called Cohen– 0 d Macaulay ﬁltered if for all i ∈{0, ... ,d} the module δ (M)/δ (M) is either zero or i i−1 Cohen–Macaulay. Notice that from the deﬁnition, it immediately follows that if δ (M)/δ (M) = 0, then it i i−1 has dimension i. A Cohen–Macaulay ﬁltered module M is sequentially Cohen–Macaulay. In fact, we may let i denote the smallest integer such that δ (M) = 0and set M = δ (M). Then, if i 1 i 1 i j 1 1 is the smallest integer such that δ (M) δ (M), we let M = δ (M). The resulting i −1 i j i j j j ﬁltration 0 = M M ... M = M is clearly a sCM ﬁltration of M. 0 1 r A ﬁltration 0 = C ⊆ C ⊆ C ⊆ ... ⊆ C = M such that each of its quotients −1 0 1 C /C is either zero or Cohen–Macaulay of dimension i is called a CM ﬁltration of M. i i−1 To see that the notions of sequentially Cohen–Macaulay and Cohen–Macaulay ﬁltered modules coincide, we need the following result, which shows that if a CM ﬁltration of a module exists, then it is unique, and it is equal to its dimension ﬁltration. Proposition 5 [[36], Proposition 4.3] Let M be a d-dimensional ﬁnitely generated R- module. If M has a CM ﬁltration 0 = C ⊆ C ⊆ C ⊆ ... ⊆ C = M, then C = δ (M) −1 0 1 d i i for every i ∈{0, ... ,d}. Proof We proceed by induction on d.If d = 0, then C = M = δ (M) and this case is 0 0 complete. Assume henceforth that d > 0, so that by induction, we have that C = δ (C ) i i d−1 Caviglia et al. Res Math Sci (2022) 9:40 Page 9 of 27 40 for all i < d. Since for i < d we have that δ (M) = δ (δ (M)), it suﬃces to show that i i d−1 δ (M) = C . d−1 d−1 Observe that Ass(C ) ⊆ Ass(M) and, for all p ∈ Ass(C /C ), we have dim A/p = i 0 0 i i−1 if C = C . With this information, we can inductively show that dim C i and, i i−1 i accordingly C ⊆ δ (M) for all i; in particular, C ⊆ δ (M). If Ass(M) =∅, then i i d−1 d−1 d−1 δ (M) = 0byRemark 3, and the desired equality is trivial. Otherwise, we let a = a = d−1 d−1 p and we claim that H (M/C ) = 0. This is obvious if M/C = 0. a d−1 d−1 p∈Ass(M) d−1 Thus, assume C = M and observe that a contains a regular element of M/C .Tosee d−1 d−1 the latter, assume that a ⊆ p; then, by prime avoidance, we can ﬁnd p ∈ p∈Ass(M/C ) d−1 Ass(M/C ) such that a ⊆ p and, therefore, p contains a prime p ∈ Ass(M) .Thisisa d−1 d−1 contradiction, since it would imply d = dim(M/C ) = dim(R/p) dim(R/p ) d −1. d−1 Finally, consider the short exact sequence 0 −→ C −→ M −→ M/C −→ 0. d−1 d−1 0 0 By Proposition 3,wehavethat δ (M) = H (M) = H (C ) ⊆ C .Thus, C = d−1 d−1 d−1 d−1 a a δ (M), and we are done. d−1 Theorem 2 (Schenzel) Let M be a ﬁnitely generated R-module; then M is sequentially Cohen–Macaulay if and only if M is Cohen–Macaulay ﬁltered. Moreover, if M is sequen- tially Cohen–Macaulay, then its sCM ﬁltration is unique. Proof We only need to show that the “only if” part. Let M :0 = M M ... M = 0 1 r M be any sCM ﬁltration of M,with d = dim(M), d =−1and d = dim(M /M ). For 0 i i i−1 j ∈{−1, 0, ... ,d},wealsolet i(j) be the largest integer i such that d j, and we set C = i j M . In this way, we have constructed a ﬁltration C :0 = C ⊆ C ⊆ ... ⊆ C = M, −1 0 i(j) d which is a CM ﬁltration of M.ByProposition 5, we may conclude that C = δ (M), and i i that C is the unique dimension ﬁltration of M.Thus, M is Cohen–Macaulay ﬁltered. Observe that one can reconstruct M from C; it follows that the modules M only depend on M as well, and M is also unique. Notice that by the above theorem and Proposition 4,if M is sequentially Cohen–Macaulay with sCM ﬁltration M, then Ass(M) = Ass(M /M ). The comparison between sCM i i i−1 ﬁltrations and dimension ﬁltrations has also the following useful consequences. Corollary 2 Let M be a d-dimensional ﬁnitely generated R-module. The following are equivalent: (1) M is sequentially Cohen–Macaulay; (2) The modules δ (M) and M/δ (M) are sequentially Cohen–Macaulay for all i ∈ i i {0, ... ,d}; (3) There exists i ∈{0, ... ,d} such that δ (M) and M/δ (M) are sequentially Cohen– i i Macaulay. In particular, M is sequentially Cohen–Macaulay if and only if M/H (M) is sequentially Cohen–Macaulay. Proof Observe that given any i ∈{0, ... ,d},wehavethat δ (M) = δ (δ (M)) for every j j i j i,and δ (M/δ (M)) = δ (M)/δ (M) for every j i. It follows that 0 ⊆ δ (M) ⊆ ... ⊆ j i j i 0 δ (M)and 0 = δ (M)/δ (M) ⊆ δ (M)/δ (M) ⊆ ... ⊆ δ (M)/δ (M)are thedimension i i i i+1 i d i ﬁltrations of δ (M)and M/δ (M), respectively. This yields all the assertions at once. i i 40 Page 10 of 27 Caviglia et al. Res Math Sci (2022) 9:40 Recall now that a ﬁnitely generated R-module M is unmixed if dim(R/p) = dim(M) for all p ∈ Ass(M). Corollary 3 Let M be a ﬁnitely generated unmixed R-module. Then, M is sequentially Cohen–Macaulay if and only if M is Cohen–Macaulay. Proof One direction is clear. Let d = dim(M); in view of Remark 3,the fact that M is unmixed guarantees that δ (M) = 0 for all i < d.ByTheorem 2, M is Cohen–Macaulay ﬁltered, and thus δ (M)/δ (M) = M is Cohen–Macaulay. d d−1 2.2 Peskine’s characterization We will always assume that R has a canonical module ω . In our setup, this assumption is not too restrictive: it is always the case when R is standard graded, and it is true for instance if R is complete local, cf. Remark 1. Proposition 6 Let R be an n-dimensional Cohen–Macaulay ring with canonical module ω and M be sequentially Cohen–Macaulay, with sCM ﬁltration 0 = M M ... R 0 1 M = M; also let d = dim M /M ,for i = 1, ... ,r. Then, r i i i−1 n−d n−d i i (1) for all i = 1, ... , r, one has that Ext (M, ω ) Ext (M /M , ω ) is Cohen– R i i−1 R R R Macaulay and has dimension d ; n−j (2) Ext (M, ω ) = 0 whenever j ∈{ / d , ... ,d }; R 1 r n−d n−d i i (3) Ext (Ext (M, ω ), ω ) M /M for i = 1, ... ,r. R R i i−1 R R Proof In the graded case, (1) and (2) follow immediately from Proposition 1,gradedlocal duality [10, Theorem 3.6.19] and [10, Theorem 3.3.10 (c) (i)]. In the local case, local duality yields (1) and (2) for the completion M of M as an R-module, see Example 2 (2). Since i i ∼ ∼ ∼ ω = ω = ω ⊗ R and Ext (N, ω ) = Ext (N, ω ) ⊗ R for any ﬁnitely generated R R R R R R R R R-module N, we conclude by faithful ﬂatness of R that (1) and (2) hold also in the local case. n−d n−d Finally, by [10, Theorem 3.3.10 (c) (iii)], we have that N = Ext (Ext (N, ω ), ω ) R R R R for any Cohen–Macaulay module N of dimension d and, thus, the last statement follows immediately from (1). Remark 4 Let M be as in the above proposition. (1) By Parts (2) and (3) of the previous proposition, we have an isomorphism n−t n−t M Ext (Ext (M, ω ), ω ), 1 R R R R where t = depth M and M is t-dimensional and Cohen–Macaulay. We will show an extensionofthisfactinLemma 1. (2) Notice that if M is sequentially Cohen–Macaulay with depth(M) = 0, then in par- ticular, M = H (M). In fact, using [10, Theorem 3.3.10 (c) (iii)] and the short exact 0 0 sequence 0 −→ H (M) −→ M −→ M/H (M) −→ 0, we get that m m n n n n 0 0 ∼ ∼ ∼ M = Ext (Ext (M, ω ), ω ) = Ext (Ext (H (M), ω ), ω ) = H (M). 1 R R R R R R R R m m We recall now the following crucial lemma, see [31, Lemma 1.5]. Caviglia et al. Res Math Sci (2022) 9:40 Page 11 of 27 40 Lemma 1 Let R be a Cohen–Macaulay n-dimensional ring with canonical module ω and M be a ﬁnitely generated R-module. n−t Let also depth(M) = t and assume that Ext (M, ω ) is Cohen–Macaulay of dimension n−t n−t t. Then, there is a natural monomorphism α :Ext (Ext (M, ω ), ω ) −→ Msuch R R R R that n−t n−t n−t n−t n−t Ext (α):Ext (M, ω ) −→ Ext (Ext (Ext (M, ω ), ω )ω ) R R R R R R R R is an isomorphism. Proof We only give a proof in the graded case; the local one is handled similarly. Without loss of generality, we may assume that R is a polynomial ring, as we show next. We write R = S/I, where S is a standard graded polynomial ring of dimension m over a ﬁeld k and I ⊆ S is homogeneous. Let m and n denote the graded maximal ideals of R and S, m−i respectively. By graded Local Duality [10, Theorem 3.6.19], we have that Ext (M, ω ) = i n−i i Hom (H (M),E (k)) and Ext (M, ω ) = Hom (H (M),E (k)). By Base Independence, S S R R R n R m i i H (M) = H (M) for all i. Thus, by Hom −⊗ adjointness and [10, Lemma 3.1.6], we obtain n m i i i ∼ ∼ Hom (H (M),E (k)) Hom (H (M), Hom (R, E (k))) Hom (H (M),E (k)). = = S S R S S R R n m m n−t m−t In particular, we have shown that Ext (M, ω ) = Ext (M, ω ) and that we may assume R S R S that R is a polynomial ring. Let now F :0 −→ F −→ ... −→ F −→ F −→ 0 be the minimal graded free • n−t 1 0 resolution of M; applying the functor Hom (−, ω ), we obtain the dual complex R R ∗ ∗ ∗ ∗ F :0 −→ F −→ ... −→ F −→ F −→ 0, • n−t 1 0 n−t ∗ ∗ with H (F ) = Ext (M, ω ). Observe that F is a complex of free modules, since ω 0 R R n−t R(−n). If we also let G denote the minimal graded free resolution of Ext (M, ω ), then • R n−t there is a map of complexes φ : F −→ G which lifts the identity map of Ext (M, ω ). • • R Since the last one is Cohen–Macaulay of dimension t, the length of G isthesameasthatof F ,namely n − t. Applying the functor Hom (−, ω )to φ , we obtain a map of complexes R R • ∗ ∗ ∗∗ φ : G −→ F which, in turn, gives a map on the zeroth cohomology: • • • ∗ n−t n−t ∗ ∗∗ α = H (φ ):Ext (Ext (M, ω ), ω ) = H (G ) → H (F ) H (F ) = M. 0 R R 0 0 0 • • R R • • Applying again the functor Hom (−, ω ), this time to φ , we obtain a map of complexes R R ∗∗ ∗∗∗ ∗∗ φ : F → G and, thus, a map • • • ∗∗ ∗∗∗ 0 ∗ n−t ∼ ∼ H (φ ): H (F ) = H (F ) = Ext (M, ω ) 0 0 R • • • n−t n−t n−t ∗∗ −→ Ext (Ext (Ext (M, ω ), ω ), ω ) = H (G ), R R R 0 R R R • n−t ∗∗ which coincides with the map Ext (α). The canonical isomorphism ψ : G −→G , n−t together with the fact that Ext (M, ω ) is Cohen–Macaulay of dimension t, gives an n−t n−t n−t n−t isomorphism H (ψ):Ext (Ext (Ext (M, ω ), ω ), ω ) −→ Ext (M, ω ), and one 0 R R R R R R R R can verify that ∗∗ H (ψ) ◦ H (φ ) = H (φ ) = id n−t . 0 0 0 • • Ext (M,ω ) Note that H (ψ) is the inverse of the isomorphism of [10, Theorem 3.3.10 (c) (iii)], and ∗∗ n−t this implies that H (φ ) = Ext (α) is the natural isomorphism of the same theorem. To conclude the proof, it remains to be shown that α is a monomorphism. 40 Page 12 of 27 Caviglia et al. Res Math Sci (2022) 9:40 n−t n−t Let N = Ext (Ext (M, ω ), ω ). We show that α is injective once we localize at every R R R R associated prime of N and then we are done, since, if ker(α) = 0, its associated primes would be contained in those of N. Let p ∈ Ass(N); since N is Cohen–Macaulay of dimension t,wehavethatdim(R/p) = t and dim(R ) = n − t. By replacing R with R , M with M ,and ω with (ω ) = ω , p p p R R p R the proof will be complete once we show that α is injective in the case t = 0. To this end observe that as in Remark 4, the short exact sequence 0 → H (M) → M → M/H (M) → 0and [10, Theorem 3.3.10 (c) (iii)] yield n n n n 0 0 ∼ ∼ Ext (Ext (M, ω ), ω ) Ext (Ext (H (M), ω ), ω ) H (M), = = R R R R R R R R m m and α composed with this isomorphism becomes just the inclusion of H (M)inside M. The equivalence between the ﬁrst two conditions in the next theorem was announced in [37] without a proof, but citing a spectral sequence argument due to Peskine. Here, we give another proof of this fact, see also [36, Theorem 5.5] where the equivalence with the third condition is proved. Theorem 3 (Peskine) Let R be a Cohen–Macaulay ring of dimension n with canonical module ω , and M be a ﬁnitely generated d-dimensional R-module. Then, the following are equivalent: (1) M is sequentially Cohen–Macaulay; n−i (2) Ext (M, ω ) is either 0 or Cohen–Macaulay of dimension i for all i ∈{0, ... ,d}; n−i (3) Ext (M, ω ) is either 0 or Cohen-Macaulay of dimension i for all i ∈{1, ... ,d − 1}. Proof The implication (1) ⇒ (2) follows at once by Proposition 6, and clearly (2) implies (3). Now assume (3); we proceed by induction on d − t, where t = depth(M). If t = d, then M is Cohen–Macaulay, and hence sequentially Cohen–Macaulay. Assume that t < d.If t = 0, then 0 = Ext (M, ω ) has ﬁnite length, and hence it is Cohen– n−t Macaulay. Either way, thanks to our assumption we have that 0 = Ext (M, ω )is t-dimensional and Cohen–Macaulay. By Lemma 1, there is an injective homomorphism n−t n−t α :Ext (Ext (M, ω ), ω ) −→ M. Since the ﬁrst module is t-dimensional Cohen– R R R R Macaulay by [10, Theorem 3.3.10 (c)], then so is its image, say M , which is a submodule of M. It follows that depth(M ) = dim(M ) = t = depth(M) < d = dim(M). 1 1 Consider the short exact sequence 0 −→ M −→ M −→ M/M −→ 0 and the 1 1 induced sequence in cohomology obtained by applying the functor Hom (−, ω ). We R R j j then have isomorphisms Ext (M/M , ω ) Ext (M, ω ) for all j = n − t, n − t + 1and 1 R = R R R the exact sequence n−t n−t n−t 0 → Ext (M/M , ω ) → Ext (M, ω ) → Ext (M , ω ) 1 R R 1 R R R R n−t+1 → Ext (M/M , ω ) → 0. 1 R n−t By Lemma 1 we know that the map Ext (α) is an isomorphism, and therefore β is an j j isomorphism as well. It follows that Ext (M, ω ) Ext (M/M , ω ) for every j = n − t, R 1 R R R n−t while Ext (M/M , ω ) = 0. 1 R This shows in particular that depth(M/M ) > t;since dim(M/M ) = d, we may apply 1 1 induction and obtain that M/M is sequentially Cohen–Macaulay. Let 0 = M /M 1 1 1 Caviglia et al. Res Math Sci (2022) 9:40 Page 13 of 27 40 M /M ... M /M = M/M be a sCM ﬁltration. Since M is Cohen–Macaulay of 2 1 r 1 1 1 dimension t and depth(M /M ) = depth(M/M ) > t by Example 1 (4) , we deduce that 2 1 1 0 = M M M ... M = M is a sCM ﬁltration and, thus M is sequentially 0 1 2 r Cohen–Macaulay. Corollary 4 Let R be Cohen–Macaulay with canonical module ω , and M, N be ﬁnitely generated R-modules. Then, M and N are sequentially Cohen–Macaulay if and only if M ⊕ N is sequentially Cohen–Macaulay. Proof Let n = dim R. We have already showed in Example 2 (1) that, if M and N are sequentially Cohen–Macaulay, then so is M ⊕ N. This also follows immediately from n−i n−i Theorem 3, since if Ext (M, ω )and Ext (N, ω )iseitherzeroorCohen–Macaulay R R R R n−i n−i n−i of dimension i,thensoisExt (M, ω ) ⊕ Ext (N, ω ) = Ext (M ⊕ N, ω ). For the R R R R R R converse, it suﬃces to observe that if the direct sum of two modules is zero or Cohen– Macaulay of a given dimension, then so is each of its summand. Remark 5 We remark that sequential Cohen–Macaulayness behaves well with respect to localization; see for instance [16, Proposition 4.7] or [15, Proposition 2.6]. For any sequentially Cohen–Macaulay R-module M and p ∈ Supp(M), one has that M is a sequentially Cohen–Macaulay R -module and, in fact, one can recover its dimension ﬁltration from that of M.Let s = dim(R/p) and consider the quotients δ (M)/δ (M) i i−1 of the dimension ﬁltration of M. If not zero, they are i-dimensional Cohen–Macaulay, and their localization is either zero or Cohen–Macaulay of dimension i − s.Now let N = (δ (M)) for all i 0 such that i + s d = dim(M), i.e., for i = 0, ... ,d − s, i i+s p and observe that 0 ⊆ N ⊆ ... ⊆ N = M is a CM ﬁltration of M , with quotients 0 d−s p p N /N (δ (M)/δ (M)) .If R is Cohen–Macaulay, the fact that M is sequentially i i−1 i+s i+s−1 p p Cohen–Macaulay for all p ∈ Supp(M) is also a consequence of Theorem 3.See [39] for other results about localization and sequentially Cohen–Macaulay modules. We now recall some deﬁnitions needed to state the next result, and that we will use frequently in the next sections. Given an R-module N,welet Ass (N) = Ass(N) {m}. Deﬁnition 4 Let M = 0 be a ﬁnitely generated graded R-module. (1) A homogeneous element 0 = y ∈ m is ﬁlter regular for M if y ∈ / ◦ p. p∈Ass (M) A sequence of homogeneous elements y , ... ,y ∈ m is a ﬁlter regular sequence for 1 t M if y is a ﬁlter regular element for M/(y , ... ,y )M for all i ∈{0, ... ,t − 1}. i+1 1 i (2) A homogeneous element 0 = y ∈ m is strictly ﬁlter regular for M if y ∈ / ◦ p, where X(M) = Ext (M, R). p∈Ass (X(M)) i∈N R A sequence of homogeneous elements y , ... ,y ∈ m is a strictly ﬁlter regular 1 t sequence for M if y is a strictly ﬁlter regular element for M/(y , ... ,y )M for i+1 1 i all i ∈{0, ... ,t − 1}. Remark 6 Regular sequences are clearly ﬁlter regular sequences. Moreover, strictly ﬁlter regular sequences are ﬁlter regular by the graded version of [9, Corollary 11.3.3]. When the ﬁeld k is inﬁnite, by Prime Avoidance any sequence of general forms is strictly ﬁlter regular. 40 Page 14 of 27 Caviglia et al. Res Math Sci (2022) 9:40 We conclude this section with the following result, which clariﬁes how sequential Cohen–Macaulayness behaves with respect to quotients, cf. Example 3;itwillalsobe useful later on. Proposition 7 Let R be a Cohen–Macaulay ring of dimension n with canonical module ω ; let M be a d-dimensional ﬁnitely generated R-module, and x ∈ R a strictly ﬁlter regular element for M. Then, (1) If M is sequentially Cohen–Macaulay, then M/xM is sequentially Cohen–Macaulay. n−i (2) The converse holds if x is regular for all nonzero Ext (M, ω ) with i > 0. Proof Assume that M is sequentially Cohen–Macaulay. Since x is strictly ﬁlter regular, n−i it is ﬁlter regular by Remark 2.17, L = 0: x has ﬁnite length and Ext (M, ω ) M R n−i Ext (M, ω ) for all i > 0, where M = M/L. Then, the long exact sequence obtained ·x by applying the functor Hom (−, ω ) to the short exact sequence 0 −→ M −→ M −→ R R M/xM −→ 0 gives a long exact sequence ·x n−(i−1) n−(i−1) n−i ... −→ Ext (M, ω ) −→ Ext (M/xM, ω ) −→ Ext (M, ω ) → ... R R R R R R ·x n−1 n n ... −→ Ext (M, ω ) −→ Ext (M/xM, ω ) −→ Ext (M, ω ) −→ 0. R R R R R R n−i By Theorem 3, we have that each nonzero Ext (M, ω ) is Cohen–Macaulay of dimension n−i i, and in this case x is Ext (M, ω )-regular when i > 0. For i > 1, we then have short exact sequences ·x n−(i−1) n−i n−i 0 −→ Ext (M, ω ) −→ Ext (M, ω ) −→ Ext (M/xM, ω ) −→ 0, R R R R R R n−(i−1) n−i and it follows that Ext (M/xM, ω ) Ext (M, ω ) ⊗ R/(x) is Cohen–Macaulay R R R R R of dimension i −1 for all i > 1. We conclude that M/xM is sequentially Cohen–Macaulay using the implication (3) ⇒ (1) of Theorem 3. n−i For the converse, the fact that x is regular for all nonzero Ext (M, ω )with i > 0 shows that the above long exact sequence of Ext modules breaks into short exact sequences ·x n−(i−1) n−i n−i 0 −→ Ext (M, ω ) −→ Ext (M, ω ) −→ Ext (M/xM, ω ) −→ 0. R R R R R R n−(i−1) for all i > 1. By Theorem 3,wehavethatExt (M/xM, ω )iseitherzeroorCohen– n−i Macaulay of dimension i − 1, and thus Ext (M, ω )iseitherzeroorCohen–Macaulay n−1 of dimension i for all i > 1. Since x is assumed to be regular on Ext (M, ω ), and n−1 n−1 dim(Ext (M, ω )) 1, we have that Ext (M, ω ) is either zero, or Cohen–Macaulay R R R R of dimension 1. It follows again from the implication (3) ⇒ (1) of Theorem 3 that M is sequentially Cohen–Macaulay. Remark 7 There are many other interesting results about sequentially Cohen–Macaulay modules and their characterizations which do not ﬁnd space in this note. For instance, in [14, Theorem 5.1], it is proven that a module is sequentially Cohen–Macaulay if and only if each module of its dimension ﬁltration is pseudo Cohen–Macaulay. In [18, Theorem 1.1], the sequential Cohen–Macaulayness of M is characterized in terms of the existence of one good system of parameters of M which has the property of parametric decomposition; see also Theorems 3.9 and 4.2 in [14] for other characterizations which involve good systems of parameters and dd-sequences. Moreover, in [15], it is investigated how the sequential Cohen–Macaulay property behaves in relation to taking associated graded rings and Rees algebras, see also [38] for more results of this type. Caviglia et al. Res Math Sci (2022) 9:40 Page 15 of 27 40 3 Partially sequentially Cohen–Macaulay modules We are going to study next the notion of partially sequentially Cohen–Macaulay module, as introduced in [35], which naturally generalizes that of sequentially Cohen–Macaulay modules. Thanks to Schenzel’s Theorem 2, the deﬁnition can be given in terms of the dimension ﬁltration of the module. Throughout this section, we let R = k[x , ... ,x ]bea 1 n standard graded polynomial ring over an inﬁnite ﬁeld k with homogeneous maximal ideal m = (x , ... ,x ). Recall that, in this case, R has a graded canonical module ω = R(−n). 1 n R We consider ﬁnitely generated graded R-modules M; when M = 0, we set depth(M) = +∞ and dim(M) =−1, as usual. We let d = dim(M). Deﬁnition 5 Let i ∈{0, ... ,d} and let {δ (M)} be the dimension ﬁltration of M; M j j is called i-partially sequentially Cohen–Macaulay, i-sCM for short, if δ (M)/δ (M)is j j−1 either zero or Cohen–Macaulay for all i j d. By deﬁnition and Corollary 2, a module M is 0-sCM if and only if M sequentially Cohen– Macaulay if and only if M is 1-sCM. Example 7 (1) Let M be a sequentially Cohen–Macaulay module. The simplest way of constructing an i-sCM module which is not sequentially Cohen–Macaulay is perhaps taking a non-sequentially Cohen–Macaulay module N of dimension strictly smaller than i, and consider their direct sum M ⊕ N,cf. Example 2 (1). (2) Let M = R/I, where I = (x ) ∩ (x ,x ) ∩ (x ,x ,x ) ⊂ R = k[x ,x ,x ,x ,x ]. With 1 2 3 4 5 1 2 3 4 5 the help of Proposition 3, we can construct the dimension ﬁltration 0 = δ = δ = −1 0 δ ⊆ δ = ((x ) ∩ (x ,x ))/I ⊆ δ = (x )/I ⊆ δ = R/I of M.Then, δ /δ and 1 2 1 2 3 3 1 4 4 3 δ /δ are Cohen–Macaulay of dimension 4 and 3, respectively, but 0 = δ /δ is not 3 2 2 1 Cohen–Macaulay. Hence, M is an example of a 3-sCM which is not 2-sCM. Remark 8 Observe that M is i-sCM if and only if M/δ (M) is sequentially Cohen– i−1 Macaulay. This follows at once recalling that the dimension ﬁltration {γ } of M/δ (M) j j i−1 is such that γ = δ (M)/δ (M) for j i and γ = 0 otherwise. Notice that, since γ = 0 j j i−1 j j for all j i − 1, if M is i-sCM then H (M/δ (M)) = 0 for all j i − 1, by Proposition m i−1 Given a graded free presentation of M F /U,wedenoteby {e , ... ,e } agradedbasis 1 r of F.Weconsider R together with the pure reverse lexicographic ordering > such that x > ... > x ; recall that > is not a monomial order on R, but by deﬁnition it agrees with 1 n the reverse lexicographic order that reﬁnes it on monomials of the same degree. We extend > to F in the following way: given monomials ue and ve of F,set i j ue > ve if deg(ue ) > deg(ve ) , or deg(ue ) = deg(ve )and u > v i j i j i j or deg(ue ) = deg(ve ),u = v and i < j . i j We shall consider this order until the end of the section, and denote by Gin(U)the generic initial module of U with respect to >. Since the action of GL (k)on R as change of coordinates can be extended in an obvious way to F,Gin(U) simply results to be the initial submodule in (gU) where g is a general change of coordinates. We prove next some preliminary facts which are needed later on. Given a graded sub- module V ⊆ F with dim(F /V ) = d, for all j ∈{−1, ... ,d},wedenoteby V the 40 Page 16 of 27 Caviglia et al. Res Math Sci (2022) 9:40 R-module such that V /V = δ (F /V ). Several results contained in the next two lemmata can be found in [22], where they are proved in the ideal case. Lemma 2 With the above notation, j j (1) Gin(U ) ⊆ Gin(U) ; j j j (2) U = (U ) ; j j (3) if V is a graded submodule of F such that U ⊆ V , then U ⊆ V ; j j j (4) Gin(U ) = Gin(U) . j j Proof (1) Notice that Gin(U )/ Gin(U)and U /U have the same Hilbert series, hence the same dimension, which is less than or equal to j.Since Gin(U) / Gin(U) = δ (F / Gin(U)), we have the desired inclusion. (2) Since U ⊆ U , one inclusion is clear. Now, consider the short exact sequence j j j j j j 0 −→ U /U −→ (U ) /U −→ (U ) /U −→ 0. j j j j Since the dimensions of U /U and (U ) /U are less than or equal to j, it follows that j j j j j also dim((U ) /U) j and, hence, (U ) ⊆ U . j j j (3) Since U ⊆ U ∩ V , we have that dim((U + V )/V ) dim(U /U) j, which j j j implies U ⊆ U + V ⊆ V . j j j (4) Since U ⊆ U , it immediately follows from (1) that Gin(U) ⊆ Gin(U ) ⊆ Gin(U) j j j and, accordingly, Gin(U) ⊆ Gin(U ) . On the other hand, by Parts (1) and (2), the j j j latter is contained in (Gin(U) ) = Gin(U) . We denote the Hilbert series of a graded R-module N by Hilb(N) = Hilb(N, z). We also let h (N) = Hilb(H (N)). Lemma 3 Let M F /U, with dimension ﬁltration {δ } ; then, the following holds: j j (1) M is i-sCM if and only if M/H (M) is i-sCM. j j j ∼ ∼ (2) If M is i-sCM, then H (M) H (δ ) H (δ /δ ) for all j i. = = m m j m j j−1 j j j (3) If M is i-sCM, then (z − 1) h (M) = (1 − z) Hilb(δ /δ ) for all j i. j j−1 (4) Let {γ } be the dimension ﬁltration of F / Gin(U); if M is i-sCM, then Hilb(δ /δ ) = j j j j−1 Hilb(γ /γ ) for all j i. j j−1 0 0 Proof Since H (R) = δ , the dimension ﬁltration of M/H (R)is {δ /δ } , which shows 0 j 0 j m m the ﬁrst part. The long exact sequence in cohomology induced by 0 → δ → δ → j−1 j δ /δ → 0 and the Cohen–Macaulayness of δ /δ for all j i easily imply (2). j j−1 j j−1 Letusﬁx j i and prove (3). If j = 0 the assertion is clear; thus we may assume j > 0 and by way of Part (2) that δ /δ is j-dimensional Cohen–Macaulay. Let x be a (δ /δ )- j j−1 j j−1 regular element of degree one; then, the short exact sequence given by multiplication by x induces a short exact sequence in cohomology δ /δ j j−1 j−1 j j 0 −→ H −→ H (δ /δ )(−1) −→ H (δ /δ ) −→ 0 m m j j−1 m j j−1 x(δ /δ ) j j−1 j−1 j together with (2) imply that h ((δ /δ )/(x(δ /δ )) = (z − 1) h (δ /δ ) = (z − j j−1 j j−1 j j−1 1) h (M). 0 j j Thus, one can easily prove that h ((δ /δ )/(x(δ /δ ))) = (z − 1) h (M), where x j j−1 j j−1 isa(δ /δ )-maximal regular sequence. Since dim((δ /δ )/(xδ /δ )) = 0, we also j j−1 j j−1 j j−1 Caviglia et al. Res Math Sci (2022) 9:40 Page 17 of 27 40 0 j have h ((δ /δ )/(xδ /δ )) = Hilb((δ /δ )/(xδ /δ )) = (1 − z) Hilb(δ /δ ), and j j−1 j j−1 j j−1 j j−1 j j−1 the proof of (3) is complete. j j−1 j j−1 Finally, to prove (4), we show that Hilb(U /U ) = Hilb(Gin(U) /(Gin(U) )) j j holds for all j i. Actually, we prove more, i.e., that Hilb(U ) = Hilb(Gin(U) ) for all j j j i;since Gin(U ) ⊆ Gin(U) by Lemma 2 (1) for all j, the last equality is equivalent j j to proving that Gin(U ) = Gin(U) for all j i, and this is what we do. j j−1 j−1 Consider now, for all j, the short exact sequences 0 → U /U → F /U → j j F /U → 0; we see inductively that depth(F /U ) j + 1 for all j i.For j = d j j−1 j j−1 and if U /U = 0, this is obvious; otherwise, since M is i-sCM, U /U is j- dimensional Cohen–Macaulay for all j i,and by [10, Proposition 1.2.9], we get that j−1 depth F /U min{j, j + 1}= j. For all graded submodules V ⊆ F,itiswell-knownthatdepth(F / Gin(V )) = depth(F /V ) and that F / Gin(V ) is sequentially Cohen–Macaulay. Thus, j + 1 depth(F / Gin(U )) by what we proved above, and Proposition 1 together with Theorem 2 imply that the latter j j t is also equal to the smallest integer t such that Gin(U ) Gin(U ) . Therefore, we j j j have shown that Gin(U ) = Gin(U ) for all j i. Now, the conclusion follows from Lemma 2 (4). Deﬁnition 6 Let M be a ﬁnitely generated graded R-module. We let δ(M) = 0if M = 0, and we let δ(M) be the largest graded R-submodule of M of dimension at most dim(M) −1 otherwise. Given j 0, we also deﬁne the module δ (M) inductively by letting 0 1 j j−1 δ (M) = M, δ (M) = δ(M), and δ (M) = δ(δ (M)). Since δ (M) is the largest submodule of M of dimension at most i, it is easy to see that for all i ∈{0, ... ,d} there exists j = j(i) 0 such that δ (M) = δ (M). In particular, δ (M) = δ(M) = δ (M). d−1 In the following remark, we collect two known facts which are useful in the following. Remark 9 Recall that in our setting R ω (n). It is awell-knownfactthatfor a d- n−i dimensional graded R-module M it holds that dim Ext (M, ω ) i for all i. (1) Given a graded submodule N ⊆ M such that dim(N) < dim(M) = d,wehave that N = δ(M)ifand only if M/N is unmixed of dimension d, i.e., dim(R/p) = dim(M/N) = d for all p ∈ Ass(M/N). This is a straightforward application of Proposition 3. (2) A d-dimensional ﬁnitely generated graded R-module M is unmixed if and only if n−j dim(Ext (M, R)) < j for all j ∈{0, ... ,d − 1}.Infact, if p ∈ Ass(M) were a prime of n−j n−j height n−j for some j < d, we would have that Ext (M, R) = Ext (M ,R ) = 0, p p p R R since the latter is, up to shift, the Matlis dual of H (M ) which is not zero because pR n−j depth(M ) = 0by[10, Proposition 1.2.13]. It follows that p ∈ Supp(Ext (M, R)) n−j and, thus, dim(Ext (M, R)) j, contradiction. The converse is analogous, observ- n−j ingthatifExt (M, R)has dimensionatleast j and, hence, necessarily equal to j, then it must have a prime of height n − j in its support. The following lemma can be regarded as an enhanced graded version of [17,Proposition 4.16]. 40 Page 18 of 27 Caviglia et al. Res Math Sci (2022) 9:40 Lemma 4 Let M be a ﬁnitely generated graded R-module. For a suﬃciently general homogeneous element x ∈ m and for all j 0 there is a short exact sequence 0 → j j+1 δ (δ(M)/xδ(M)) → δ (M/xM) → L → 0, where L is a module of ﬁnite length. j j Proof Let d = dim(M). The case d 1 is trivial; therefore, we will assume that d 2 and proceed by induction on j 0. First assume that j = 0; let M = M/δ(M), and observe that M is unmixed of posi- tive depth and dimension d. In particular, Ext (M, R) = 0and,byRemark 9,wehave n− dim(Ext (M, R)) < for all 0 << d. For x suﬃciently general, we have that x is M-regular and, thus, δ(M) ∩xM = x(δ(M): x) = xδ(M); moreover, either dim(δ(M)/xδ(M)) 0ordim(δ(M)/xδ(M)) = dim(δ(M))− 1 < d − 1 = dim(M/xM) = dim(M/xM). If we let T = δ(M) + xM, then δ(M)/xδ(M) = T /xM ⊆ δ(M/xM), and therefore, we have an exact sequence 0 → δ(M)/xδ(M) → ∼ ∼ δ(M/xM). Moreover, we have that coker(ϕ) = δ((M/xM)/(T /xM)) = δ(M/T). Since M/T = M/xM, we then have a short exact sequence 0 −→ δ(M)/xδ(M) −→ δ(M/xM) −→ δ(M/xM) −→ 0. Since x is suﬃciently general, by Remark 6, we may assume that x is also strictly ﬁlter regu- n− n− lar for M. We then have that either dim(Ext (M, R) ⊗ R/(x)) = dim(Ext (M, R)) 0 R R n− n− or dim(Ext (M, R) ⊗ R/(x)) = dim(Ext (M, R)) − 1 < − 1 for all 0 << d.Since R R x is strictly ﬁlter regular for M, from the short exact sequences n−(−1) n− 0 −→ Ext (M, R) ⊗ R/(x) −→ Ext (M/xM, R) R R −→ 0: x −→ 0 n−(−1) Ext (M,R) n−(−1) n− it also follows that dim(Ext (M/xM, R)) = dim(Ext (M, R) ⊗ R/(x)) < − 1 for R R all 0 − 1 < d − 1. Thus, by Remark 9, it follows that M/xM is unmixed of dimension d − 1 and that L = δ(M/xM) is necessarily δ (M/xM) = H (M/xM), which has ﬁnite 0 0 length. Now suppose that the statement of the lemma is proved for j −1, so that we have a short j−1 j exact sequence 0 → δ (δ(M)/xδ(M)) → δ (M/xM) → L → 0, with L of ﬁnite j−1 j−1 length. It is clear from the deﬁnition of δ that there is an exact sequence 0 −→ j j+1 δ (δ(M)/xδ(M)) −→ U −→ L −→ 0, where we let U = δ (M/xM)and L = coker(ϕ ). j j j If U has ﬁnite length we are done, so let us assume that dim(U) > 0. In this case, we nec- essarily have that h = dim(δ (M/xM)) > 0, and since L has ﬁnite length we conclude j−1 that dim(δ(M)/xδ(M)) = h. Again because L has ﬁnite length, we can ﬁnd p 0 such j−1 p j−1 p p that m U ⊆ δ (δ(M)/xδ(M)). Moreover, dim(m U) dim(U) < h, and therefore m U j p is contained in δ (δ(M)/xδ(M)). This shows that m L = 0, and thus L has ﬁnite length. j j In [35], it is claimed that if x ∈ R is M-regular, it is possible to prove that M is i-sCM if and only if M/xM is (i − 1)-sCM following the same lines of the proof [36, Theorem 4.7], which is not utterly correct, as we already pointed out in Example 3. The claim is indeed false: if x is M-regular and M/xM is (i − 1)-sCM, then M is not necessarily i-sCM, as the following example shows. Example 8 Let R be a 2-dimensional domain which is not Cohen–Macaulay, cf. Example 3.For all0 = x ∈ R, we have that R/(x)is0-sCM,but R is not even 2-sCM. Caviglia et al. Res Math Sci (2022) 9:40 Page 19 of 27 40 In the following proposition, we show that the claimed result holds true under some additional assumption. Proposition 8 Let i be a positive integer, and M be a ﬁnitely generated graded R-module n− with depth(M) > 0. Also assume that depth(Ext (M, R)) > 0 for all i − 1.For a suﬃciently general x ∈ R, if M/xM is (i − 1)-sCM, then M is i-sCM. Proof Since a module is 0-sCM if and only if it is 1-sCM, cf. Deﬁnition 5 and Corollary 2, when i = 1, 2 the statement follows immediately from Proposition 7; therefore, we may let 3 i d = dim(M). Let j be the smallest integer such that δ (M) = δ (M), and let i−1 N = M/δ (M). Observe that depth(N) > 0byProposition 4.ByRemark 8,itisenoughto prove that N is sequentially Cohen–Macaulay, and this is what we do. n− n− Observe that since dim(δ (M)) i − 1, we have that Ext (M, R) = Ext (N, R) for R R n−(i−1) n−(i−1) all i, and there is an injection 0 → Ext (N, R) → Ext (M, R). In particular, R R from our assumptions, we get that n− depth(Ext (N, R)) > 0 for all i − 1. By a repeated application of Lemma 4, there exists a short exact sequence j j j 0 −→ δ (M)/xδ (M) −→ δ (M/xM) −→ L −→ 0, j j j where L is a module of ﬁnite length. Now, either dim(δ (M/xM)) = dim(δ (M)/xδ (M)) = j j j dim(δ (M)) − 1, or dim(δ (M/xM)) 0, and in both cases, we have that dim(δ (M/xM)) j−1 i − 2. By minimality of j, we have that dim(δ (M)) > i − 1 and applying iteratively j−1 j−1 Lemma 4 as we did above, we also obtain dim(δ (M/xM)) = dim(δ (M)) − 1 > i − 2; thus, we may conclude that δ (M/xM) = δ (M/xM). i−2 j j j Since L = coker(ϕ)and x is regular for N = M/δ (M), we have (M/xM)/(δ (M)/xδ (M)) ∼ ∼ M/(δ (M) + xM) N /xN, and the above yields a short exact sequence 0 → L → = = N /xN → (M/xM)/δ (M/xM) → 0. i−2 Since L has ﬁnite length, the associated long exact sequence of Ext-modules yields that n− n− Ext (N /xN, R) Ext ((M/xM)/δ (M/xM),R) for all = 0. i−2 R R In particular, being M/xM a(i − 1)-sCM module by assumption, Remark 8 and Peskine n− Theorem 3 imply that Ext (N /xN, R) is either zero or Cohen–Macaulay of dimension for all = 0. Remark 8 together with local duality also imply that n−(−1) Ext (N /xN, R) = 0 for all 1 − 1 i − 2. n− We may assume that x, which is N-regular, is also Ext (N, R)-regular for all i − 1, for we proved above that all these modules have positive depth. For i, we thus have ·x n−(−1) n− n− short exact sequences 0 → Ext (N, R) −→ Ext (N, R) → Ext (N /xN, R) → 0, R R R n− from which it follows that Ext (N, R) is either zero or Cohen–Macaulay of dimension for all i. ·x n− From the above, we also have that for all 2 i − 1 the maps Ext (N, R) −→ n− n− Ext (N, R) are isomorphisms, and by graded Nakayama’s Lemma that Ext (N, R) = 0 R R ·x n−1 n−1 for all 2 i−1. Finally, we also have an injection 0 → Ext (N, R) −→ Ext (N, R), R R n−1 which implies that Ext (N, R) is Cohen–Macaulay of dimension one by Remark 9. Applying Peskine’s Theorem 2.9, we have thus showed that N is sequentially Cohen– Macaulay, that is, M is i-sCM. 40 Page 20 of 27 Caviglia et al. Res Math Sci (2022) 9:40 The next theorem provides a characterization of partially sequentially Cohen–Macaulay modules. It was proved for the ﬁrst time in [35, Theorem 3.5] in the ideal case. Here, we generalize the result to ﬁnitely generated modules and ﬁx the gap in the original proof thanks to Proposition 8.Welet R = k[x , ... ,x ] = R/x R and denote by N the [n−1] 1 n−1 n [n−1] R -module N /x N N ⊗ R/x R by restriction of scalars. We let x ∈ R be a general [n−1] n R n linear form which, without loss of generality, we may write as l = a x +···+a x −x 1 1 n−1 n−1 n and consider the map g : R → R ,deﬁnedby x → x for i = 1, ... ,n − 1and n [n−1] i i x → a x + ··· + a x . Then, the surjective homomorphism F /U → F /g (U) n 1 1 n−1 n−1 n [n−1] has kernel (U + xF)/U and induces the isomorphism F F [n−1] = . (1) U + xF g (U) Moreover, the image of Gin(U)in F via the mapping x → 0isGin(U) .With [n−1] [n−1] this notation, the module version of [24, Corollary 2.15] states that Gin(g (U)) = Gin(U) . (2) [n−1] Theorem 4 Let M be a ﬁnitely generated graded R-module of dimension d, and let M F /U be a free graded presentation of M. The following conditions are equivalent: (1) F /Uis i-sCM; j j (2) h (F /U) = h (F / Gin(U)) for all i j d. Proof (1) ⇒ (2) is a direct consequence of Lemma 3 (3) and (4), since also F / Gin(U)is i-sCM. We prove the converse by induction on d.If d = 0, F /U is Cohen–Macaulay and sequentially Cohen–Macaulay. Therefore, without loss of generality we may assume that F /U and F / Gin(U) have positive dimension and, by Lemma 3 (1), also positive depth. Since F / Gin(U) is sequentially Cohen–Macaulay, Peskine’s Theorem 3 implies that there exists a linear form l ∈ R which is F / Gin(U)-regular and also regular n−j for all non-zero Ext (F / Gin(U), ω )with j > 0. Starting with the exact sequence ·l 0 → F / Gin(U)(−1) → F / Gin(U) → F /(Gin(U) + lF) → 0, by the above and local duality, we obtain the short exact sequences j−1 j 0 −→ H (F /(Gin(U) + lF)) −→ H (F / Gin(U))(−1) m m ·l −→ H (F / Gin(U)) −→ 0, j−1 j from which it follows that h (F /(Gin(U) + lF)) = (z − 1)h (F / Gin(U)) for all j. Consider now a suﬃciently general linear form x ∈ R.For all j, there are exact sequences j−1 j j (j) (j) 0 −→ B −→ H (F /(U + xF)) −→ H (F /U)(−1) −→ H (F /U) −→ C −→ 0 m m m (j) (j) j−1 j for some R-modules B and C , and these imply that h (F /(U +xF)) = (z−1)h (F /U)+ (j) (j) Hilb(B ) + Hilb(C ) for all j.By (1)and (2), we obtain j j (j) (j) (z − 1)h (F /U) (z − 1)h (F /U) + Hilb(B ) + Hilb(C ) j−1 j−1 = h (F /(U + xF)) = h (F /g (U)) [n−1] n j−1 j−1 h (F / Gin(g (U))) = h (F / Gin(U) ) [n−1] n [n−1] [n−1] j−1 j = h (F /(Gin(U) + lF)) = (z − 1)h (F / Gin(U)). Thus, from our hypothesis, it follows that the above inequalities are equalities for all j i, (j) (j) (j) (j) that means that Hilb(B ) = Hilb(C ) = 0, i.e., B = C = 0 for j i. Moreover, since Caviglia et al. Res Math Sci (2022) 9:40 Page 21 of 27 40 n−j (i−1) (i) C = B = 0, it follows that x is regular for all nonzero Ext (F /U, ω )with j i −1, which thus have positive depth. j j From the above equalities we also get that h (F /g (U)) = h (F / Gin(g (U))) for n n [n−1] [n−1] all j i − 1; by induction, this implies that F /(U + xF) F /g (U)is(i − 1)-sCM. [n−1] n The conclusion follows now by a straightforward application of Proposition 8. As a corollary, we immediately obtain Theorem 1. Theorem 5 Let M be a ﬁnitely generated graded R-module, and let M F /Ua free graded presentation of M. Then, F /U is sequentially Cohen–Macaulay if and only if i i h (F /U) = h (F / Gin(U)) for all j 0. i i One can wonder whether the equality h (F /U) = h (F / Gin(U)) is enough to imply that F /U is i-sCM; however, this is not the case. Example 9 (1) Consider a graded Cohen–Macaulay k[x , ... ,x ]-module M of dimen- 1 n 1 sion i and a graded non-sequentially Cohen–Macaulay k[x , ... ,x ]-module N. i+2 n Let M = N ⊗ k[x , ... ,x ] and take M = M ⊕ M .Then, x , ... ,x 2 k 1 i+1 1 2 1 i+1 is a strictly ﬁlter-regular sequence for M , and it follows from Proposition 7 (1) that M is not sequentially Cohen–Macaulay. By Corollary 4 also M is not sequentially Cohen–Macaulay. On the other hand, since depth(M ) > i,wehave i i ∼ ∼ ∼ that H (M) = H (M ). If we write M = F /U and M = F /U , where 1 1 1 1 2 2 2 m m F and F are graded free R-modules and U ,U are graded submodules, then 1 2 1 2 M ⊕ M = F /U where F = F ⊕ F and U = U ⊕ U , and it follows that 1 2 1 2 1 2 Gin(U) = Gin(U ) ⊕ Gin(U ). Since M is Cohen–Macaulay, hence sequentially 1 2 1 Cohen–Macaulay, and depth(F / Gin(U )) = depth(M ) > i, we therefore conclude 2 2 2 i i i i by Theorem 5 that h (F /U) = h (F /U ) = h (F / Gin(U )) = h (F / Gin(U)). 1 1 1 1 (2) The following is another explicit example of such instance in the ideal case. Consider the polynomial ring R = k[x ,x ,x ,x ,x ,x ,x ] and the monomial ideal 1 2 3 4 5 6 7 3 2 2 2 3 I = (x ,x x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x ). Then, 2 4 1 5 1 6 2 5 2 6 3 5 3 6 3 7 4 5 4 6 4 7 1 1 2 7 7 one can check that depth(R/I) = 0and dim(R/I) = 3; moreover, h (R/I) = j j j h (R/ Gin(I)) for j = 0, 3, and h (R/I) = h (R/ Gin(I)) for j = 1, 2. By Theorem 4, this means that R/I is 3-sCM but not 2-sCM and, a fortiori, not 1-sCM. lex On the other hand, if I is the lexicographic ideal associated with I, then the equality i i lex h (R/I) = h (R/I ) ensures the i-partial sequential Cohen–Macaulayness of R/I.Notice i i i lex that this is a stronger condition, though, since h (R/I) h (R/ Gin(I)) h (R/I ), coef- ﬁcientwise, see [34, Theorems 2.4 and 5.4]. Actually, in [35, Theorem 4.4], the following result is proved. Theorem 6 Let i be a positive integer and I a homogeneous ideal of R; then, the following conditions are equivalent: i i lex (1) h (R/I) = h (R/I ); i i lex (2) h (R/ Gin(I)) = h (R/I ); j j lex (3) h (R/I) = h (R/I ) for all j i. If any of the above holds, then I is i-sCM. 40 Page 22 of 27 Caviglia et al. Res Math Sci (2022) 9:40 We conclude this section by observing that the conditions in the previous theorem are still equivalent if we replace Gin(I) with Gin (I), the zero-generic initial ideal of I introduced in [13]. We also remark that the equivalence between conditions (2) and (3) is not true if we replace the Hilbert series of local cohomology modules with graded Betti numbers, see [32, Theorem 3.1]. 4E-depth As in the previous section, k will denote an inﬁnite ﬁeld, R = k[x , ... ,x ]astandard 1 n graded polynomial ring and m = (x , ... ,x ) its homogeneous maximal ideal. As before, 1 n when M = 0, we let dim(M) =−1and depth(M) =∞. We start with the main deﬁnition of this section. Deﬁnition 7 Given a nonzero ﬁnitely generated graded Z-module M and a nonnegative integer r, we say that M satisﬁes condition (E ) if depth(Ext (M, R)) min{r, n − i} for all i ∈ Z.Welet E-depth(M) = min n, sup{r ∈ N | M satisﬁes (E )} . Remark 10 Observe that a nonzero module M is sequentially Cohen–Macaulay if and only if it satisﬁes condition (E ) for all r 0, see Theorem 3. It is therefore clear that a sequentially Cohen–Macaulay R-module has E-depth equal to n. The converse is also i i true, since if E-depth(M) = n, then depth(Ext (M, R)) n −i;asdim(Ext (M, R)) n −i R R always holds, cf. Remark 9, this implies the claim. Lemma 5 Let M = 0 be a ﬁnitely generated Z-graded R-module with positive depth and E-depth, and let be a linear form which is a strictly ﬁlter regular for M; then, the graded short exact sequence 0 −→ M(−1) −→ M −→ M/M −→ 0 induces graded short exact sequences i i i+1 0 −→ Ext (M, R) −→ Ext (M, R)(1) −→ Ext (M/M, R) −→ 0,foralli < n, R R R and i−1 i i 0 −→ H (M/M) −→ H (M)(−1) −→ H (M) −→ 0,foralli > 0. m m m Proof Consider the induced long exact sequence of Ext modules i i i+1 ... Ext (M, R) Ext (M(−1),R) Ext (M/M, R) ... , R R R i i and observe that Ext (M(−1),R) = Ext (M, R)(1). Our assumption that E-depth(M) > 0 R R guarantees that depth(Ext (M, R)) > 0 for all i < n.Since is strictly ﬁlter regular for M, the multiplication by is injective on Ext (M, R) for all i < n, and the long exact sequence breaks into short exact sequences, as claimed, and the graded short exact sequences of local cohomology modules are obtained by graded Local Duality. Proposition 9 Let M = 0 be a ﬁnitely generated Z-graded R-module. (1) E-depth(M) = E-depth(M/H (M)). (2) Assume that E-depth(M) > 0 and is homogeneous strictly ﬁlter regular for M; then, 0 0 either E-depth(M/(H (M) + M)) = E-depth(M) = n, or E-depth(M/(H (M) + m m M)) = E-depth(M) − 1. Caviglia et al. Res Math Sci (2022) 9:40 Page 23 of 27 40 Proof We ﬁrst prove (1); clearly, we may assume that H (M) = 0. Applying the functor 0 0 Hom (−,R) to the short exact sequence 0 −→ H (M) −→ M −→ M/H (M) −→ 0 m m n−i we get a long exact sequence of Ext modules, from which we obtain that Ext (M, R) n−i 0 0 Ext (M/H (M),R) for all i = 0 because H (M) has ﬁnite length; the ﬁrst statement is m m n 0 now clear, since depth(Ext (H (M),R)) = 0. For the proof of (2), let N = M/H (M); by (1), M and N have same E-depth and i i Ext (M, R) = Ext (N, R) for all i = n. First assume that M is sequentially Cohen– R R Macaulay, i.e., E-depth(M) = n.Then, N is sequentially Cohen–Macaulay by Remark 10 or by Corollary 2, and since is ﬁlter regular for M by Remark 2.17, is N-regular. It follows from Proposition 7 that M/(H (M) + M) is sequentially Cohen–Macaulay and, therefore, has E-depth equal to n. Now assume that 0 < E-depth(N) = E-depth(M) = r < n; from an application i i of Lemma 5, we obtain short exact sequences 0 → Ext (N, R) → Ext (N, R)(1) → R R i+1 i+1 i i Ext (N /N, R) → 0 and, therefore, Ext (N /N, R) = Ext (M, R)(1)/ Ext (M, R) for R R R R i+1 all i < n. It follows that depth(Ext (N /N, R)) = depth(Ext (M, R)) − 1 for all i < n, R R which clearly implies E-depth(N /N) = E-depth(M) − 1, as desired. We now introduce a special grading on a polynomial ring which reﬁnes the standard grading, and that can be further reﬁned to the monomial Z -grading. Deﬁnition 8 Let r be a positive integer, A be a Z-graded ring and S = A[y , ... ,y ]a 1 r r+1 polynomial ring over A.For i ∈{0, ... ,r} let η ∈ Z be the vector whose (i + 1)-st entry is 1, and all other entries equal 0. We consider S as a Z × Z -graded ring by letting deg (a) = deg (a) · η , for all a ∈ A and deg (y ) = η , for i ∈{1, ... ,r}. 0 i i S A S By means of the previous deﬁnition, we may consider R = k[x , ... ,x ]asa Z × Z -graded 1 n ring for any 0 r n−1 by letting deg (x ) = η for all 1 i n−r and deg (x ) = η for i 0 i i R R all n −r +1 i n. Observe that an element f ∈ R is graded with respect to such grading if and only if f can be written as f = f · u, where f ∈ k[x , ... ,x ] is homogeneous with 1 n−r respect to the standard grading, and u is a monomial in k[x , ... ,x ]. In particular, n−r+1 n when r = 0 this is just the standard grading on R, while for r = n − 1, it coincides with the monomial Z -grading. Remark 11 We can extend in a natural way such a grading to free R-modules F with a basis by assigning degrees to the elements of the given basis. Accordingly, any R-module r r M is Z × Z -graded if and only if M F /U, where F is a free Z × Z -graded R-module and U is a Z × Z -graded submodule of F. We see next that the grading just introduced is very relevant to the purpose of estimating the E-depth of a module; the following can be seen as a reﬁnement of Proposition 2. Proposition 10 Let R = k[x , ... ,x ], and M be a ﬁnitely generated Z × Z -graded R- 1 n module such that x , ... ,x is a ﬁlter regular sequence for M. Then, x , ... ,x is n n−r+1 n n−r+1 a strictly ﬁlter regular sequence for M, and E-depth(M) r. Proof Write M as M = F /U, where F is a ﬁnitely generated Z × Z -graded free R- r r module, and U is a Z × Z -graded submodule of F.Since U is Z × Z -graded, x is r sat ∞ r Z × Z -homogeneous and ﬁlter regular, U = U : x is also Z × Z graded and, n 40 Page 24 of 27 Caviglia et al. Res Math Sci (2022) 9:40 0 sat sat accordingly M/H (M) F /U is too. If F /U = 0, then M has dimension zero and, thus, is sequentially Cohen–Macaulay; then E-depth(M) = n t, and every sequence of non-zero elements of (x , ... ,x ) is a strictly ﬁlter regular sequence for M. In particular, 1 n x , ... ,x is a strictly ﬁlter regular sequence for M. n n−r+1 sat Now suppose that F /U = 0. By assumption, x is ﬁlter regular for M, and thus regular sat r sat for F /U . Since the latter is Z×Z -graded,wehavethat F /U = F /U ⊗ k[x ] for some r−1 r−1 Z × Z -graded R = k[x , ... ,x ]-module F, and some Z × Z -graded submodule 1 n−1 sat U of F such that F /U can be identiﬁed with the hyperplane section F /U ⊗ R/(x ). R n Thus, for i < n,wehave i i 0 i sat i ∼ ∼ ∼ Ext (M, R) Ext (M/H (M),R) Ext (F /U ,R) Ext (F /U, R) ⊗ k[x ]. = = = k n R R m R It follows that x is a nonzero divisor on Ext (M, R) for all i < n, and thus E-depth(M) > 0. Since Ext (F /U, R) has ﬁnite length it also follows that x is a strictly ﬁlter regular element for M. Now we can consider F /U, and an iteration of this argument will imply the desired conclusion. We now introduce a weight order on R. Given integers 0 r n, consider the following r × n matrix ⎡ ⎤ 00 ... 00 0 0 ... 0 −1 ⎢ ⎥ ⎢ 00 ... 00 0 0 ... −10 ⎥ ⎢ ⎥ . . . . . . . . ⎢ . ⎥ . . . . . . . . . ⎢ . ⎥ . . . . . . . . = ⎢ ⎥ , r,n ⎢ ⎥ 00 ... 00 0 −1 ... 00 ⎢ ⎥ ⎢ ⎥ 00 ... 00 −10 ... 00 ⎣ ⎦ 00 ... 0 −10 0 ... 00 and let ω be its i-th row; then, this induces a “partial” revlex order rev on R by declaring i r a a b b a 1 n b 1 n that a monomial x = x ··· x is greater than a monomial x = x ··· x if and only n n 1 1 if there exists 1 j r such that ω · a = ω · b for all i j and ω · a >ω · b. i i j+1 j+1 Remark 12 Observe that rev coincides with the usual revlex order on R = k[x , ... ,x ]. n 1 n Give a graded free R-module F with basis {e , ... ,e }, we can write any f ∈ F uniquely 1 s as a ﬁnite sum f = u e , where u are monomials, and we assume the sum has minimal j i j support. We let the initial form in (f )of f written as above be the sum of those u e for rev j i r j which u is maximal with respect to the order rev introduced above. j r Deﬁnition 9 Let F be a ﬁnitely generated graded free R-module, and U be a graded submodule; let also M = F /U. We say that the r-partial general initial submodule of U satisﬁes a given property P if there exists a non-empty Zariski open set L of r-uples of linear forms such that F /in (g (U)) satisﬁes P for any = ( , ... , ) ∈ L, where rev n−r+1 n g is the automorphism on F induced by the change of coordinates of R which sends → x and ﬁxes the other variables. i i With some abuse of notation, we denote any partial initial submodule in (g (U)) rev which satisﬁes P the r-partial general submodule of U, and denote it by gin (U). Proposition 11 Let F be a ﬁnitely generated graded free R-module, and U ⊆ Fbe a graded submodule. For every 0 r n, we have that E-depth(F / gin (U)) r. r Caviglia et al. Res Math Sci (2022) 9:40 Page 25 of 27 40 Proof For a suﬃciently general change of coordinates, we have that x , ... ,x form n n−r+1 a ﬁlter regular sequence for F /(g (U)). Since in (U): x = in (U : x ), see [21, rev F n rev F n r r 15.7], we have that x , ... ,x also form a ﬁlter regular sequence for F /in (g (U)), n n−r+1 rev and hence for F / gin (U). By construction, the module F / gin (U)is Z × Z -graded, and r r the claim now follows from Proposition 10. Theorem 7 Let F be a ﬁnitely generated graded free R-module, and U ⊆ F beagraded submodule. For every 0 r n, we have that E-depth(F /U) rif and only if h (F /U) = h (F / gin (U)) for all i ∈ N. Proof After performing a suﬃciently general change of coordinates, we may assume that V = in (U) has the same properties as gin (U), and that x , ... ,x form a strictly rev n n−r+1 r rev ﬁlter regular sequence for F /U. By using again that in (U : x ) = in (U): x , rev F n rev F n r r and because strictly ﬁlter regular sequences are ﬁlter regular by Remark 6,wehavethat x , ... ,x form a ﬁlter regular sequence for F /V.Since V is Z × Z -graded, it follows n n−r+1 from Proposition 10 that x , ... ,x is a strictly ﬁlter regular sequence for F /V.By n n−r+1 [21, 15.7], we also have that sat ∞ ∞ sat V = in (U): x = in (U : x ) = in (U ), rev F rev F rev r n r n r and sat sat sat V + x F = in (U ) + x F = in (U + x F). n rev n rev n r r sat Viewing F /(U + x F) as a quotient of a free S = k[x , ... ,x ]-module F by a graded n 1 n−1 sat submodule U,wesee that V + x F canbeidentiﬁedwithasubmodule V ⊆ F,with V = in (U)and V has the same properties of gin (U), see [12, Lemma 3.4] and rev r−1 rev t−1 the proof of [12, Theorem 3.6] for more details. First assume that E-depth(F /U) r, and we prove equality for the Hilbert series of i i H (F /U)and H (F /V ) for all i ∈ N by induction on r 0. The base case is trivial since m m sat in (U) = U.ByProposition 9,wehavethatE-depth(F /U ) r > 0 and, accordingly, rev sat i sat E-depth(F /(U + x F)) r − 1. By induction, we have that H (F /(U + x F)) = n n i sat sat sat H (F /(V + x F)). Since x is strictly ﬁlter regular for both F /U and F /V ,and n n these have positive E-depth, by Lemma 5, we have graded short exact sequences for all i > 0: ·x i−1 sat i sat i sat 0 H (F /(U + x F)) H (F /U )(−1) H (F /U ) 0, m m m (3) and ·x i−1 sat i sat i sat 0 H (F /(V + x F)) H (F /V )(−1) H (F /V ) 0. m m m (4) Since the ﬁrst modules in both sequences have the same Hilbert series, a straightforward i sat i sat i sat i computation shows that h (F /U ) = h (F /V ). Since H (F /U ) H (F /U)and m m i sat i H (F /V ) H (F /V ) for all i > 0, the equality between Hilbert series is proved for m m sat sat i > 0. Finally, since Hilb(F /U) = Hilb(F /V ) and Hilb(F /U ) = Hilb(F /V ), we have that 0 sat sat 0 h (F /U) = Hilb(U /U) = Hilb(V /V ) = h (F /V ). 40 Page 26 of 27 Caviglia et al. Res Math Sci (2022) 9:40 Conversely, assume that the local cohomology modules of F /U and F /V have the same Hilbert series. If r = 0 there is nothing to show, otherwise it is enough to prove that sat E-depth(F /U ) r. Since x is a ﬁlter regular element for F /V , it is strictly ﬁlter regular for F /V , being the r sat latter Z × Z -graded, and hence it is ﬁlter regular for F /V . Thus, the sequence (4)is exact for i > 0. On the other hand, suppose by way of contradiction that the sequence (3) is not exact for some i ∈ Z,and let i be the smallest such integer, so that we still have an exact sequence ·x i−1 sat i sat i sat 0 H (F /(U + x F)) H (F /U )(−1) H (F /U ). m m m Counting dimensions in such a sequence, and comparing them to those obtained from i−1 sat i−1 sat (4) we obtain that h (F /(U + x F)) > h (F /(V + x F)) . However, by upper n j n j semi-continuity, we know that the reverse inequality always holds, which gives a con- tradiction. Thus, the sequence (3) is exact for all i ∈ N. By induction we have that sat sat E-depth(F /(U + x F)) r − 1. If E-depth(F /(U + x F)) = n,byProposition 9, n n we see that E-depth(F /U) = n r, as desired. Otherwise, again by Proposition 9,we sat have that E-depth(F /U) = E-depth(F /(U + x F)) + 1 r. Recalling Remark 12,Theorem 7 can be viewed as another extension of Theorem 1. Funding Open access funding provided by Università di Pisa within the CRUI-CARE Agreement. The ﬁrst author was partially supported by a grant from the Simons Foundation (41000748, G.C.). The second author was partially supported by the PRIN 2020 project 2020355B8Y “Squarefree Gröbner degenerations, special varieties and related topics”. The third author was partially supported by the PRIN project 2017SSNZAW004 “Moduli Theory and Birational Classiﬁcation” of Italian MIUR. Author details Department of Mathematics, Purdue University, 150 N. University Street, West Lafayette, IN 47907-2067, USA, 2 3 Dipartimento di Matematica, Università di Genova, Via Dodecaneso 35, 16146 Genova, Italy, Dipartimento di Matematica, Università degli Studi di Pisa, Largo Bruno Pontecorvo 5, 56127 Pisa, Italy, Dipartimento di Matematica “Giuseppe Peano”, Università degli Studi di Torino, Via Carlo Alberto 10, 10123 Torino, Italy. Received: 15 January 2022 Accepted: 20 April 2022 References 1. 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Research in the Mathematical Sciences – Springer Journals
Published: Sep 1, 2022
Keywords: Sequentially Cohen-Macaulay module; Dimension filtration; Generic initial ideals; Local cohomology
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