1 Introduction

Sequentially Cohen-Macaulay modules were first introduced in the graded setting by Stanley [20], in connection with the theory of Stanley-Reisner rings. Later, Schenzel [19] introduced the notion of Cohen-Macaulay filtered module in a more general setting, and showed that in the graded case it coincides with the one introduced by Stanley.

Let k be a field, and \(S=k[x_1,\ldots ,x_n]\) with the standard grading. Let M be a finitely generated \(\mathbb {Z}\)-graded S-module of Krull dimension d. A module M is sequentially Cohen-Macaulay if one of the following three equivalent conditions is satisfied:

  1. (1)

    (Stanley) There exists a filtration \(0 = M_0 \subsetneq M_1 \subsetneq M_2 \subsetneq \cdots \subsetneq M_t = M\) with \(M_i/M_{i-1}\) Cohen-Macaulay graded S-modules of dimension \(d_i\) satisfying \(d_i>d_{i-1}\) for all \(i=1,\ldots ,t\).

  2. (2)

    (Schenzel) If \(\delta _i(M)\) denotes the largest submodule of M of dimension less than or equal to i, then the filtration

    $$\begin{aligned} 0 \subseteq \delta _0(M) \subseteq \delta _1(M) \subseteq \cdots \subseteq \delta _{d}(M) = M \end{aligned}$$

    is such that \(\delta _i(M)/\delta _{i-1}(M)\) is either zero or Cohen-Macaulay for all \(i=1,\ldots ,d\).

  3. (3)

    (Peskine) For every \(i=0,\ldots ,n\), the module \({\text {Ext}}^{n-i}_S(M,S)\) is either zero or Cohen-Macaulay of dimension i.

Over the years, sequentially Cohen-Macaulay modules have been studied extensively, even from very different perspectives (for instance, see [1, 2, 7,8,9, 11,12,13,14]). We invite the interested reader to consult [6] for further details.

For the purposes of this note, it is important to mention a result of Herzog and Sbarra, which characterizes sequentially Cohen-Macaulay modules. We state it here in a dual version:

Theorem 1.1

([15, Main theorem]). Let F be a finitely generated free S-module, and \(U \subseteq F\) be a homogeneous submodule. Then, the module F/U is sequentially Cohen-Macaulay if and only if for all \(i = 0,\ldots ,n\), we have \({\text {HF}}({\text {Ext}}^{n-i}_S(F/U,S)) = {\text {HF}}({\text {Ext}}^{n-i}_S(F/{\text {gin}}_{revlex} (U),S))\), where \({\text {HF}}(-)\) denotes the Hilbert function of a graded S-module.

This theorem has been extended in two different directions: weaker versions of sequential Cohen-Macaulayness can be characterized either by showing that equality of Hilbert functions holds only for certain cohomological indices [17], or by replacing revlex with certain partial revlex orders [5].

To state our main theorem, we recall the notion of arithmetic degree, introduced by Bayer and Mumford in [3] (see also [21, 22]). First, recall that if \(0\ne M\) is a finitely generated \(\mathbb {Z}\)-graded module of dimension d, and \({\text {HF}}(M;j) = \dim _k(M_j)\) is the Hilbert function of M in degree j, then for \(j \gg 0\), we have

$$\begin{aligned} \sum _{i \leqslant j} {\text {HF}}(M;i) = \frac{{\text {e}}(M)}{d!} j^{d} + O(j^{d-1}) \end{aligned}$$

for some \({\text {e}}(M) \in \mathbb {Z}_{>0}\), called multiplicity (or degree) of M. Given an integer \(r \geqslant d\), we let

$$\begin{aligned} {\text {e}}_r(M) = {\left\{ \begin{array}{ll} {\text {e}}(M) &{} \text { if } r=d, \\ 0 &{} \text { otherwise. } \end{array}\right. } \end{aligned}$$

Definition 1.2

Let \(S=k[x_1,\ldots ,x_n]\), and M be a finitely generated graded S-module. For all \(r=0,\ldots ,n\), we let \({\text {adeg}}_r(M) = {\text {e}}_r({\text {Ext}}^{n-r}_S(M,S))\). The arithmetic degree of M is defined as \({\text {adeg}}(M) = \sum _{r=0}^n {\text {adeg}}_r(M)\).

Recall that \(\dim ({\text {Ext}}_S^{n-r}(M,S)) \leqslant r\) always holds (see for instance [18, Section 3.1]), therefore the above definition makes sense. Now let F be a free S-module, and \(U \subseteq F\) be a graded submodule. Given any weight \(\omega \in \mathbb {Z}^n\), by upper semi-continuity, one has \({\text {adeg}}_r(F/U) \leqslant {\text {adeg}}_r(F/{\text {in}}_\omega (U))\) for all \(r=0,\ldots ,n\) (see [21] for the case of cyclic modules); in particular, \({\text {adeg}}(F/U) \leqslant {\text {adeg}}(F/{\text {gin}}_\textrm{revlex}(U))\) always holds. If F/U is sequentially Cohen-Macaulay, then Theorem 1.1 immediately gives that \({\text {adeg}}(F/U) = {\text {adeg}}(F/{\text {gin}}_\textrm{revlex}(U))\). This result was observed by Lu and Yu [16, Proposition 3.6], and in the same paper they conjecture that the converse holds as well. The purpose of this note is to prove their conjecture:

Main theorem

Let \(S=k[x_1,\ldots ,x_n]\), with the standard grading. Let F be a finitely generated graded free S-module and \(U \subseteq F\) be a homogeneous submodule. We have that F/U is sequentially Cohen-Macaulay if and only if \({\text {adeg}}(F/U) = {\text {adeg}}(F/{\text {gin}}_\textrm{revlex}(U))\).

Actually, we prove their conjecture in a bigger generality, by showing that it is not necessary to take general coordinates when computing initial ideals as long as \(x_n,\ldots ,x_1\) is a filter regular sequence for F/U (see Sect. 2). As a consequence of our proof, we also obtain an analogous generalization of Theorem 1.1.

2 Preliminaries and main result

Let k be a field, and \(S=k[x_1,\ldots ,x_n]\) be a polynomial ring with standard grading \(\deg (x_i)=1\) for all i. Let \(\mathfrak {m}=(x_1,\ldots ,x_n)\). Throughout, M will always denote a finitely generated \(\mathbb {Z}\)-graded S-module.

Consider the weight \(\omega =(0,\ldots ,0,-1) \in \mathbb {Z}^n\) and, for a homogeneous submodule U of a graded free S-module F, we let \({\text {in}}(U):= {\text {in}}_\omega (U) \subseteq F\) be the initial submodule. Since revlex can be obtained as \({\text {in}}_\Omega \) for the following matrix of weights

$$\begin{aligned} \Omega = \begin{bmatrix} 0 &{} \quad 0 &{}\quad \ldots &{} \quad 0&{}\quad -1 \\ 0 &{} &{}\quad 0 \ldots &{}\quad -1 &{}\quad -1 \\ \vdots &{} \quad \vdots &{} \quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad -1 &{} \quad \ldots &{}\quad -1 &{}\quad -1 \end{bmatrix}, \end{aligned}$$

of which \(\omega \) is the first row, we call \({\text {in}}(U)\) a “partial revlex submodule”. See [5] for more details on this construction, where \({\text {in}}(-)\) is denoted as \({\text {in}}_{\textrm{rev}_{1}}(-)\).

Definition 2.1

Let M be a finitely generated \(\mathbb {Z}\)-graded S-module. A homogeneous element \(f \in S\) is called filter regular if \(0:_M f\) has finite length. A sequence of elements \(f_1,\ldots ,f_t\) is called a filter regular sequence if \(f_{i+1}\) is filter regular for \(M/(f_1,\ldots ,f_i)M\) for all \(i<t\).

Remark 2.2

Let U be a homogeneous submodule of a graded free S-module F. Since \({\text {in}}(U)\) is a partial deformation towards \({\text {in}}_\textrm{revlex}(U)\), one has that \({\text {adeg}}_r(F/U) \leqslant {\text {adeg}}_r(F/{\text {in}}(U)) \leqslant {\text {adeg}}_r(F/{\text {in}}_\textrm{revlex}(U))\) for all \(r=0,\ldots ,n\) by upper semicontinuity.

In the proof of our main theorem, we will need the following lemma, which is an immediate consequence of a result of Serre.

Lemma 2.3

Let M be a finitely generated \(\mathbb {Z}\)-graded S-module of dimension \(d>0\), and let \(\ell \in S_1\) be such that \(M/\ell M\) is Cohen-Macaulay of dimension \(d-1\), and \({\text {e}}_{d-1}(M/\ell M) = {\text {e}}_d(M)\). Then \(\ell \) is an M-regular element, and M is Cohen-Macaulay.

Proof

Without loss of generality, we may assume that k is infinite. Then, we can find linear forms \(\ell _2,\ldots ,\ell _{d}\) which form a regular sequence for \(M/\ell M\). Since \(M/\ell M\) is Cohen-Macaulay, our assumptions guarantee that \({\text {e}}_d(M) = {\text {e}}_{d-1}(M/\ell M) = \lambda (M/(\ell ,\ell _2,\ldots ,\ell _d)M)\), where \(\lambda (-)\) denotes the length of a module. By the graded version of [4, Theorem 4.7.10(b)], we conclude that M is Cohen-Macaulay and \(\ell ,\ell _2,\ldots ,\ell _d\) is a regular sequence on M. \(\square \)

For a module M, we let \({\text {Ass}}_r(M) = {\text {Ass}}(M) \cap \{\mathfrak {p}\in {\text {Spec}}(S) \mid \dim (S/\mathfrak {p}) = r\}\).

Remark 2.4

We recall that, if M is finitely generated and \(\mathbb {Z}\)-graded, then \({\text {Ass}}_r(M) = {\text {Ass}}_r({\text {Ext}}^{n-r}_S(M,S))\) for all \(r = 0,\ldots ,n\). This follows by noticing that, for a prime \(\mathfrak {p}\) such that \(\dim (S/\mathfrak {p}) = r\), we have that \(\mathfrak {p}\in {\text {Ass}}_r(M)\) if and only if \(H^0_{\mathfrak {p}S_\mathfrak {p}}(M_\mathfrak {p}) \ne 0\), if and only if \(\left( {\text {Ext}}^{n-r}_S(M,S)\right) _\mathfrak {p}\ne 0\). Since \(\dim ({\text {Ext}}^{n-r}_S(M,S))\leqslant r\) (see for instance [18, Section 3.1]), this is equivalent to \(\mathfrak {p}\in {\text {Min}}({\text {Supp}}({\text {Ext}}^{n-r}_S(M,S)))\), which in turn is equivalent to \(\mathfrak {p}\in {\text {Ass}}_r({\text {Ext}}^{n-r}_S(M,S))\).

Remark 2.5

Let F be a finitely generated free S-module, and \(U \subseteq F\) be a homogeneous submodule. By [15, Theorem 2.2], we have that \(F/{\text {gin}}_\textrm{revlex}(U)\) is sequentially Cohen-Macaulay. However, the same argument works for \(F/{\text {in}}_\textrm{revlex}(U)\) assuming that \(x_n,x_{n-1},\ldots ,x_1\) form a filter regular sequence for F/U (see also [5, Key example 2.15]). In particular, under these assumptions, for all \(r=0,\ldots ,n\), the module \({\text {Ext}}^{n-r}_S(F/{\text {in}}_\textrm{revlex}(U),S)\) is either zero or Cohen-Macaulay of dimension r.

We are now ready to prove the conjecture of Lu and Yu [16].

Theorem 2.6

Let \(S=k[x_1,\ldots ,x_n]\), with the standard grading. Let F be a finitely generated graded free S-module and \(U \subseteq F\) be a homogeneous submodule. Assume that \(x_n,x_{n-1},\ldots ,x_1\) is a filter regular sequence for F/U. The following are equivalent:

  1. (1)

    F/U is sequentially Cohen-Macaulay,

  2. (2)

    \({\text {HF}}({\text {Ext}}^{n-r}_S(F/U,S)) = {\text {HF}}({\text {Ext}}^{n-r}_S(F/{\text {in}}_\textrm{revlex}(U),S))\) for all \(r=0,\ldots ,n\),

  3. (3)

    \({\text {adeg}}(F/U) = {\text {adeg}}(F/{\text {in}}_\textrm{revlex}(U))\).

Proof

We prove the equivalence of the three conditions by induction on \(d=\dim (F/U)\). Let \(V={\text {in}}_\textrm{revlex}(U)\). First of all, observe that \(U^{{\text {{sat}}}} = U:x_n^\infty \), and that \({\text {in}}_\textrm{revlex}(U^{{\text {{sat}}}}) = V:x_n^\infty = V^{{\text {{sat}}}}\) by well-known properties of revlex-type orders (see [10, 15.7]). It follows that \({\text {HF}}(U^{{\text {{sat}}}}/U) = {\text {HF}}(V^{{\text {{sat}}}}/V)\). Since \(U^{{\text {{sat}}}}/U\) is the graded Matlis dual of \({\text {Ext}}^n_S(F/U,S)\) (and similarly for V), we conclude that \({\text {HF}}({\text {Ext}}^n_S(F/U,S)) = {\text {HF}}({\text {Ext}}^n_S(F/V,S))\). In particular, \({\text {adeg}}_0(F/U) = {\text {e}}_0({\text {Ext}}^n_S(F/U,S)) = {\text {e}}_0({\text {Ext}}^n_S(F/V,S)) = {\text {adeg}}_0(F/V)\).

Since modules of dimension zero are automatically sequentially Cohen-Macaulay, for \(d=0\), the three statements are trivially equivalent because they are all true.

Now assume that \(d>0\). Recall that F/U is sequentially Cohen-Macaulay if and only if so is \(F/U^{{\text {{sat}}}}\) [6, Corollary 2.8], and that \({\text {Ext}}^{n-r}_S(F/U,S) \cong {\text {Ext}}^{n-r}_S(F/U^{{\text {{sat}}}},S)\) for all \(r>0\). Similar considerations hold for V. In view of the previous equalities, we may assume that \({\text {depth}}(F/U)>0\) after possibly replacing U with \(U^{{\text {{sat}}}}\).

Assume (1). By Remark 2.4, we have \({\text {Ass}}_r(F/U)= {\text {Ass}}_r({\text {Ext}}^{n-r}_S(F/U,S))\). Also, \({\text {Ass}}_r({\text {Ext}}^{n-r}_S(F/U,S)) = {\text {Ass}}({\text {Ext}}^{n-r}_S(F/U,S))\) as \({\text {Ext}}^{n-r}_S(F/U,S)\) is either zero or Cohen-Macaulay of dimension r. Since \(x_n\) is filter regular for F/U, we conclude that it is regular for \({\text {Ext}}^{n-r}_S(F/U,S)\) for all \(r>0\). For all \(r>0\), we have short exact sequences

which give that \(F/(U+x_nF)\) is sequentially Cohen-Macaulay of dimension \(d-1\). Moreover, for all \(r>0\) and \(j \in \mathbb {Z}\), we obtain

$$\begin{aligned}{} & {} {\text{ HF }}({\text{ Ext }}^{n-r+1}_S(F/(U+x_nF),S);j)\nonumber \\ {}{} & {} \quad = {\text{ HF }}({\text{ Ext }}^{n-r}_S(F/U,S);j+1)- {\text{ HF }}({\text{ Ext }}^{n-r}_S(F/U,S);j). \end{aligned}$$
(2.1)

Now let \(\overline{S} = k[x_1,\ldots ,x_{n-1}]\). We can identify \(F/(U+x_nF)\) with a sequentially Cohen-Macaulay \(\overline{S}\)-module \(\overline{F}/\overline{U}\), where \(\overline{F}\) is a graded free \(\overline{S}\)-module and \(\overline{U} \subseteq \overline{F}\) is a homogeneous submodule. Since \(\dim (\overline{F}/\overline{U}) = d-1\), by induction, we have that \({\text {HF}}\left( {\text {Ext}}^{n-r}_{\overline{S}}(\overline{F}/\overline{U},\overline{S})\right) = {\text {HF}}\left( {\text {Ext}}^{n-r}_{\overline{S}}(\overline{F}/{\text {in}}_\textrm{revlex}(\overline{U}),\overline{S})\right) \) for all \(r > 0\). By [4, Lemma 3.1.16], for all \(r>0\), we have that

$$\begin{aligned} {\text {Ext}}^{n-r}_{\overline{S}}(\overline{F}/\overline{U},\overline{S})\cong {\text {Ext}}^{n-r+1}_S(F/(U+x_nF),S) \end{aligned}$$

and

$$\begin{aligned} {\text {Ext}}^{n-r}_{\overline{S}}(\overline{F}/{\text {in}}_\textrm{revlex}(\overline{U}),\overline{S})\cong {\text {Ext}}^{n-r+1}_S(F/(V+x_nF),S). \end{aligned}$$

Using that \({\text {in}}_\textrm{revlex}(U+x_nF) = V+x_nF\) (see [10, 15.7]), we obtain

$$\begin{aligned}{} & {} {\text {HF}}({\text {Ext}}^{n-r+1}_S(F/(U+x_nF),S)) \\{} & {} \quad = {\text {HF}}({\text {Ext}}^{n-r+1}_S(F/{\text {in}}_\textrm{revlex}(U+x_nF),S)) \text { for all } r>0. \end{aligned}$$

By the proof of [5, Proposition 3.5] and by [5, Lemma 2.9], for all \(r>0\), we have short exact sequences

(2.2)

which give

$$\begin{aligned}{} & {} {\text {HF}}({\text {Ext}}^{n-r+1}_S(F/V+x_nF,S);j) \nonumber \\{} & {} \quad = {\text {HF}}({\text {Ext}}^{n-r}_S(F/VS);j+1) - {\text {HF}}({\text {Ext}}^{n-r}_S(F/V,S);j). \end{aligned}$$
(2.3)

Combining (2.1) and (2.3), together with the fact that the \({\text {Ext}}\) modules vanish for sufficiently negative degrees, we finally obtain

$$\begin{aligned} {\text {HF}}({\text {Ext}}^{n-r}_S(F/U,S)) = {\text {HF}}({\text {Ext}}^{n-r}_S(F/V,S)) \text { for all } r>0, \end{aligned}$$

and this concludes the proof that (1) \(\Rightarrow \) (2). The fact that (2) \(\Rightarrow \) (3) is trivial.

We finally show that (3) \(\Rightarrow \) (1). Let \(\omega =(0,\ldots ,0,-1) \in \mathbb {Z}^n\), and let \(W={\text {in}}_\omega (U)\). By Remark 2.2 and our assumptions, we must have \({\text {adeg}}_r(F/U)={\text {adeg}}_r(F/W) = {\text {adeg}}_r(F/V)\) for all \(r=0,\ldots ,n\). Again using the proof of [5, Proposition 3.5] and [5, Lemma 2.9], for all \(r>0\), we obtain short exact sequences

(2.4)

where we use that \(W+x_nF = U+x_nF\) holds because \(\omega \) is a partial revlex order. The short exact sequences (2.4) and (2.2) give that \({\text {adeg}}_{r-1}(F/(U+x_nF)) = {\text {adeg}}_r(F/W) = {\text {adeg}}_r(F/V) = {\text {adeg}}_{r-1}(F/(V+x_nF))\). We therefore obtain that \({\text {adeg}}(F/(U+x_nF)) = {\text {adeg}}(F/(V+x_nF))\). As above, we can identify \(F/(U+x_nF)\) with an \(\overline{S}\)-module \(\overline{F}/\overline{U}\). In this way, \(F/(V+x_nF)\) can be identified with \(\overline{F}/{\text {in}}_\textrm{revlex}(\overline{U})\), and therefore we have that \({\text {adeg}}(\overline{F}/\overline{U}) = {\text {adeg}}(\overline{F}/{\text {in}}_\textrm{revlex}(\overline{U}))\). Since \(\dim (\overline{F}/\overline{U})=d-1\) and \(x_{n-1},\ldots ,x_1\) form a filter regular sequence for \(\overline{F}/\overline{U}\), by induction, we have that \(\overline{F}/\overline{U}\) is sequentially Cohen-Macaulay, and so is \(F/(U+x_nF)\).

Now we note that if \({\text {Ext}}^{n-r}_S(F/U,S) \ne 0\), then \({\text {Ext}}^{n-r}_S(F/V,S) \ne 0\) by upper semi-continuity. Since we have \({\text {e}}_r({\text {Ext}}^{n-r}_S(F/U,S)) = {\text {adeg}}_r(F/U) = {\text {adeg}}_r(F/V) = {\text {e}}_r({\text {Ext}}^{n-r}_S(F/V,S))\), and the latter is positive by Remark 2.5, it follows that \(\dim ({\text {Ext}}^{n-r}_S(F/U,S))=r\). To complete the proof, we show by induction on \(r>0\) that if \({\text {Ext}}^{n-r}_S(F/U,S) \ne 0\), then it is a Cohen-Macaulay module, and that \(x_n\) is a regular element for it.

First note that for \(r>0\), since \({\text {Ass}}_r(F/U) = {\text {Ass}}_r({\text {Ext}}^{n-r}_S(F/U,S))\) by Remark 2.4, and because \(x_n\) is filter regular for F/U, we have that \(x_n\) avoids all minimal primes of \({\text {Supp}}({\text {Ext}}^{n-r}_S(F/U,S))\) of dimension r. In particular, \(\dim ({\text {Ext}}^{n-r}_S(F/U,S) \otimes _S S/(x_n)) = r-1\).

For \(r=1\), we have an exact sequence

which gives that \({\text {Ext}}^n_S(F/(U+x_nF),S) \cong {\text {Ext}}^{n-1}_S(F/U,S) \otimes _S S/(x_n)\). Using the short exact sequence (2.4), and recalling that \(U+x_nF = W + x_nF\), we obtain

$$\begin{aligned} {\text {e}}_1({\text {Ext}}^{n-1}_S(F/U,S))&= {\text {e}}_1({\text {Ext}}^{n-1}_S(F/W,S)) \\&= {\text {e}}_0({\text {Ext}}^n_S(F/(U+x_nF),S))\\&= {\text {e}}_0({\text {Ext}}^{n-1}_S(F/U,S) \otimes _S S/(x_n)). \end{aligned}$$

By Lemma 2.3, we conclude that \({\text {Ext}}^{n-1}_S(F/U,S)\) is Cohen-Macaulay, and \(x_n\) is regular for it.

Now assume that \({\text {Ext}}^{n-i}_S(F/U,S)\) is Cohen-Macaulay for all \(i=1,\ldots ,r-1\) and that \(x_n\) is a regular element for all such modules. In particular, it is regular for \({\text {Ext}}^{n-r+1}_S(F/U,S)\), and therefore we have an exact sequence

We conclude that \({\text {Ext}}^{n-r+1}_S(F/(U+x_nF),S) \cong {\text {Ext}}^{n-r}(F/U,S) \otimes _S S/(x_n)\). As before, we have

$$\begin{aligned} {\text {e}}_r({\text {Ext}}^{n-r}_S(F/U,S))&= {\text {e}}_r({\text {Ext}}^{n-r}_S(F/W,S)) \\&= {\text {e}}_{r-1}({\text {Ext}}^{n-r+1}_S(F/(U+x_nF),S)) \\&= {\text {e}}_{r-1}({\text {Ext}}^{n-r}_S(F/U,S) \otimes _S S/(x_n)). \end{aligned}$$

Finally, by Lemma 2.3, we conclude that \({\text {Ext}}^{n-r}_S(F/U,S)\) is Cohen-Macaulay and \(x_n\) is regular for it, and the proof is complete. \(\square \)

Remark 2.7

We would like to point out that, if the field k is infinite, then the assumption that \(x_n,\ldots ,x_1\) form a filter regular sequence on F/U is not at all restrictive. In fact, by prime avoidance, we can always find a sequence of linearly independent linear forms \(\ell _n,\ldots ,\ell _1\) which form a filter regular sequence on F/U. After performing the change of coordinates \(x_i:= \ell _i\) for \(i=1,\ldots ,n\), we are in the assumptions of Theorem 2.6.