Sequentially Cohen-Macaulay matroidal ideals

TL;DR

This paper investigates the class of sequentially Cohen-Macaulay matroidal ideals in polynomial rings, providing classifications for degree 2 and some cases for degree 3 or higher.

Contribution

It offers a complete classification of degree 2 sequentially Cohen-Macaulay matroidal ideals and partial classifications for higher degrees in specific cases.

Findings

All degree 2 sequentially Cohen-Macaulay matroidal ideals are classified.

Partial classification results for degree d ≥ 3 in certain cases.

Provides structural insights into matroidal ideals with Cohen-Macaulay properties.

Abstract

Let $R=K[x_1,...,x_n]$ be the polynomial ring in $n$ variables over a field $K$ and let $J$ be a matroidal ideal of degree $d$ in $R$. In this paper, we study the class of sequentially Cohen-Macaulay matroidal ideals. In particular, all sequentially Cohen-Macaulay matroidal ideals of degree $2$ are classified. Furthermore, we give a classification of sequentially Cohen-Macaulay matroidal ideals of degree $d\geq 3$ in some special cases.

Equations

$$\dim (M_1/M_0)<\dim (M_2/M_1)<...<\dim (M_r/M_{r-1}).$$

$$\mathrm{Ass}(R/I)=\{(x_1,x_3,x_4),(x_1,x_2,x_5),(x_2,x_3,x_4,x_5)\}.$$

$$0⟶\frac{I}{(I_i)}⟶\frac{R}{(I_i)}⟶\frac{R}{I}⟶0.$$

$$\mathrm{reg}(\frac{R}{(I_i)})=\max \{\mathrm{reg}(\frac{R}{I}),\mathrm{reg}(\frac{I}{(I_i)})\}=\max \{d-1,i-1\}=i-1.$$

$$0⟶R/(y_1{J}^{\prime \vee }:{\mathfrak{p}}^{\vee })(-n+1)\stackrel{y_2{y}_3...y_n}{\to }R/y_1{J}^{\prime \vee }⟶R/({\mathfrak{p}}^{\vee },y_1{J}^{\prime \vee })⟶0.$$

$$\mathrm{reg}(R/({\mathfrak{p}}^{\vee },y_1{J}^{\prime \vee }))$$

$$=\max \{n-2,\mathrm{reg}(R/y_1{J}^{\prime \vee })\}.$$

$$J=$$

$$y_1{y}_3...y_{d-1}{J}_{d-2}+y_2{y}_3...y_{d-1}{J}_{d-1}+J_d,$$

$$J^{\prime }=y_1{y}_2...y_{d-2}{J}_1+y_1{y}_2...y_{d-3}{y}_{d-1}{J}_2+...+y_1{y}_3...y_{d-1}{J}_{d-2}+y_2{y}_3...y_{d-1}{J}_{d-1}+J_d$$

$$(J:y_1{y}_2...y_{d-i-1}{y}_{d-i+1}...y_{d-1})=y_{d-i}\mathfrak{p}+J_i,$$

$$(J:y_1{y}_2...y_{d-i-1}{y}_{d-i+1}...y_{d-1})=y_{d-i}\mathfrak{p}+J_i$$

$$J=y_1...y_{d-1}\mathfrak{p}+y_1{y}_2...y_{d-2}{J}_1+y_1{y}_2...y_{d-3}{y}_{d-1}{J}_2+...+y_2{y}_3...y_{d-1}{J}_{d-1}+J_d,$$

$$J=y_1...y_{d-1}\mathfrak{p}+y_1{y}_2...y_{d-2}{y}_d\mathfrak{q}+y_1{y}_2...y_{d-3}{y}_{d-1}{J}_2+...+y_1{y}_3...y_{d-2}{J}_{d-2}+y_2{y}_3...y_{d-1}{J}_{d-1}+J_d,$$

$$J=y_1{y}_2\mathfrak{p}+y_1{y}_3{\mathfrak{q}}_1+y_2{y}_4{\mathfrak{q}}_2+J_1,$$

$$J=(y_1,y_4)\cap (y_2,y_3)\cap (y_1,y_2,J_1)\cap (y_2,y_3,J_1)\cap (y_1,{\mathfrak{q}}_2)\cap (y_2,{\mathfrak{q}}_1)\cap \mathfrak{p}.$$

$$J=(y_1,y_4)\cap (y_2,y_3)\cap (y_1,y_2,J_1)\cap (y_1,{\mathfrak{q}}_2)\cap (y_2,{\mathfrak{q}}_1)\cap \mathfrak{p}$$

$$J=$$

$$y_1{y}_3...y_{n-3}{J}_{n-4}+y_2{y}_3...y_{n-3}{J}_{n-3}+J_{n-2}$$

$$J=$$

$$\cap ...\cap (y_{n-4},y_{n-3})\cap (y_1,J_{n-3})\cap ...\cap (y_{n-3},J_1)$$

$$J^{\vee }=(y_1{y}_2,...,y_1{y}_{n-3},y_2{y}_3,...,y_2{y}_{n-3},...,y_{n-4}{y}_{n-3},y_1{J}_{n-3}^{\vee },...,y_{n-3}{J}_1^{\vee },{\mathfrak{p}}^{\vee }).$$

$$J=$$

$$y_1{y}_3...y_{n-3}{J}_{n-4}+y_2{y}_3...y_{n-3}{J}_{n-3}+J_{n-2}$$

$$(J:y_1)=y_2{y}_3...y_{n-3}\mathfrak{p}+y_2...y_{n-4}{J}_1+y_2...y_{n-5}{y}_{n-3}{J}_2+...+y_3...y_{n-3}{J}_{n-4}+(J_{n-2}:y_1),$$

$$(J:y_3)=$$

$$=$$

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Introduction

Our goal is to classify the sequentially Cohen-Macaulay matroidal ideals. While for the Cohen-Macaulay property of matroidal ideals, a complete classification was given by Herzog and Hibi [10], the classification of the sequentially Cohen-Macaulay matroidal ideals seems to be much harder. In the present paper partial answers to this problem are given. Herzog and Hibi [9] were the first to give a systematic treatment of polymatroidal ideals and they studied some combinatoric and algebraic properties related to it. They defined the polymatroidal ideal, a monomial ideal having the exchange property. A square-free polymatroidal ideal is called a matroidal ideal. Herzog and Takayama [13] proved that all polymatroidal ideals have linear quotients which implies that they have linear resolutions. Herzog and Hibi [10] proved that a polymatroidal ideal $I$ is Cohen-Macaulay (i.e. CM) if and only if $I$ is a principal ideal, a Veronese ideal, or a square-free Veronese ideal.

Let $R=K[x_{1},\dots,x_{n}]$ be the polynomial ring in $n$ indeterminate over a field $K$ and $I\subset R$ be a homogeneous ideal. For a positive integer $i$, let $(I_{i})$ be the ideal generated by all forms in $I$ of degree $i$. We say that I is componentwise linear if for each positive integer $i$, $(I_{i})$ has a linear resolution. Componentwise linear ideals were first introduced by Herzog and Hibi [8] to generalize Eagon and Reiner’s result that the Stanley-Reisner ideal $I_{\Delta}$ of simplicial complex $\Delta$ has a linear resolution if and only if the Alexander dual $\Delta^{\vee}$ is C.M [5]. In particular, Herzog and Hibi [8] and Herzog, Reiner, and Welker [12] showed that the Stanley-Reisner ideal $I_{\Delta}$ is componentwise linear if and only if $\Delta^{\vee}$ is sequentially Cohen-Macaulay(i.e. SCM).

It is of interest to understand the SCM matroidal ideals, and this paper may be considered as a first attempt to characterize such ideals for matroidal ideals in low degree or in a small number of variables. The remainder of this paper is organized as follows. Section 1 and 2 recall some definitions and results of componentwise linear ideals, simplicial complexes, and polymatroidal ideals. Section 3 classifies all SCM matroidal ideals of degree $2$. Section 4 studies SCM matroidal ideals of degree $d\geq 3$ over polynomial rings of small dimensional.

For any unexplained notion or terminology, we refer the reader to [11] and [21]. Several explicit examples were performed with help of the computer algebra systems Macaulay2 [7].

1. Preliminaries

In this section, we recall some definitions and results used throughout the paper. As in the introduction, let $K$ be a field and $R=K[x_{1},...,x_{n}]$ be the polynomial ring in $n$ variables over $K$ with each $\deg x_{i}=1$. Let $I\subset R$ be a monomial ideal and $G(I)$ be its unique minimal set of monomial generators of $I$.

We say that a monomial ideal $I$ with $G(I)=\{u_{1},...,u_{r}\}$ has linear quotients if there is an ordering $\deg(u_{1})\leq\deg(u_{2})\leq...\leq\deg(u_{r})$ such that for each $2\leq i\leq r$ the colon ideal $(u_{1},...,u_{i-1}):u_{i}$ is generated by a subset $\{x_{1},...,x_{n}\}$.

It is known that if a monomial ideal $I$ generated in single degree has linear quotients, then $I$ has a linear resolution (see [3, Lemma 4.1]). In particular, a monomial ideal $I$ generated in degree $d$ has a linear resolution if and only if the Castelnuovo-Mumford regularity of $I$ is $\operatorname{reg}(I)=d$ (see [20, Lemma 49]).

Lemma 1.1**.**

[4, Corollary 20.19]** If $0\longrightarrow A\longrightarrow B\longrightarrow C\longrightarrow 0$ is a short exact sequence of graded finitely generated $R$-modules, then

• (a)

$\operatorname{reg}A\leq\max(\operatorname{reg}B,\operatorname{reg}C+1).$**

• (b)

$\operatorname{reg}B\leq\max(\operatorname{reg}A,\operatorname{reg}C),$**

• (c)

$\operatorname{reg}C\leq\max(\operatorname{reg}A-1,\operatorname{reg}B).$**

• (d)

If $A$ has a finite length, set $s(A)=\max\{s:A_{s}\neq 0\}$, then $\operatorname{reg}(A)=s(A)$ and the equality holds in $(b)$.

One of the important classes of monomial ideals with linear quotients is the class of polymatroid ideals.

Let $I\subset R$ be a monomial ideal generated in one degree. We say that $I$ is polymatroidal if the following ”exchange condition” is satisfied: For any two monomials $u=x_{1}^{a_{1}}x_{2}^{a_{2}}\dots x_{n}^{a_{n}}$ and $v=x_{1}^{b_{1}}x_{2}^{b_{2}}\dots x_{n}^{b_{n}}$ belong to $G(I)$ such that $\deg_{x_{i}}(v)<\deg_{x_{i}}(u)$, there exists an index $j$ with $\deg_{x_{j}}(u)<\deg_{x_{j}}(v)$ such that $x_{j}(u/x_{i})\in G(I)$. The polymatroidal ideal $I$ is called matroidal if $I$ is generated by square-free monomials. Note that if $I$ is a matroidal ideal of degree $d$, then $\operatorname{depth}(R/I)=d-1$ (see [2]).

Theorem 1.2**.**

[10, Theorem 4.2]** A polymatroidal ideal $I$ is CM if and only if $I$ is a principal ideal, a Veronese ideal, or a square-free Veronese ideal.

2. review on componentwise linear ideals

For a homogeneous ideal $I$, we write $(I_{i})$ to denote the ideal generated by the degree $i$ elements of $I$. Note that $(I_{i})$ is different from $I_{i}$, the vector space of all degree $i$ elements of $I$. Herzog and Hibi introduced the following definition in [8].

Definition 2.1**.**

A monomial ideal $I$ is componentwise linear if $(I_{i})$ has a linear resolution for all $i$.

A number of familiar classes of ideals are componentwise linear. For example, all ideals with linear resolutions, all stable ideals, all square-free strongly stable ideals are componentwise linear (see [11]).

Proposition 2.2**.**

[6, Proposition 2.6]** If $I$is a homogeneous ideal with linear quotients, then $I$ is componentwise linear.

If $I$ is generated by square-free monomials, then we denote by $I_{[i]}$ the ideal generated by the square-free monomials of degree $i$ of $I$.

Theorem 2.3**.**

[8, Proposition 1.5]** Let $I$ be a monomial ideal generated by square-free monomials. Then $I$ is componentwise linear if and only if $I_{[i]}$ has a linear resolution for all $i$.

The notion of componentwise linearity is intimately related to the concept of sequential Cohen-Macaulayness.

Definition 2.4**.**

[18]** A graded $R$-module $M$ is called sequentially Cohen-Macaulay (SCM) if there exists a finite filtration of graded $R$-modules $0=M_{0}\subset M_{1}\subset...\subset M_{r}=M$ such that each $M_{i}/M_{i-1}$ is Cohen-Macaulay, and the Krull dimensions of the quotients are increasing:

[TABLE]

The theorem connecting sequentially Cohen-Macaulayness to componentwise linearity is based on the idea of Alexander duality. We recall the definition of Alexander duality for square-free monomial ideals and then state the fundamental result of Herzog and Hibi [8] and Herzog, Reiner, and Welker [12].

Let $\Delta$ be a simplicial complex on the vertex set $V=\{x_{1},x_{2},...,x_{n}\}$, i.e., $\Delta$ is a collection of subsets $V$ such that $(1)$ $\{x_{i}\}\in\Delta$ for each $i=1,2,...,n$ and $(2)$ if $F\in\Delta$ and $G\subseteq F$, then $G\in\Delta$. Let $\Delta^{\vee}$ denote the dual simplicial complex of $\Delta$, that is to say, $\Delta^{\vee}=\{V\setminus F\mid F\notin\Delta\}$.

If $I$ is a square-free monomial ideal, then the square-free Alexander dual of $I=(x_{1,1}...x_{1,n_{1}},...,x_{t,1}...x_{t,n_{t}})$ is the ideal $I^{\vee}=(x_{1,1},...,x_{1,n_{1}})\cap...\cap(x_{t,1},...,x_{t,n_{t}}).$

We quote the following results which are proved in [5], [8], [19] and [15].

Theorem 2.5**.**

Let $I$ be a square-free monomial ideal of $R$. Then the following conditions hold:

• (a)

$R/I$* is CM if and only if the Alexander dual $I^{\vee}$ has a linear resolution.*

• (b)

$R/I$* is SCM if and only if the Alexander dual $I^{\vee}$ is componentwise linear.*

• (c)

$\operatorname{proj\,dim}(R/I)=\operatorname{reg}(I^{\vee})$.

• (d)

If $y_{1},...,y_{r}$ is an $R$-sequence with $\deg(y_{i})=d_{i}$ and $I=(y_{1},...,y_{r})$, then $\operatorname{reg}(I)=d_{1}+...+d_{r}-r+1$.

In the following if $G(I)=\{u_{1},...,u_{t}\}$, then we set $\operatorname{supp}(I)=\cup_{i=1}^{t}\operatorname{supp}(u_{i})$, where $\operatorname{supp}(u)=\{x_{i}:u=x_{1}^{a_{1}}...x_{n}^{a_{n}},a_{i}\neq 0\}$. Also we set $\gcd(I)=\gcd(u_{1},...,u_{m})$ and $\deg(I)=\max\{\deg(u_{1}),...,\deg(u_{m})\}.$

Throughout this paper we assume that all matroidal ideals are full supported, that is, $\operatorname{supp}(I)=\{x_{1},...,x_{n}\}$.

Corollary 2.6**.**

[6, Corollary 6.6]** Let $\Delta$ be a simplicial complex on $n$ vertices, and let $I_{\Delta}$ be it’s Stanley-Reisner ideal, minimally generated by square-free monomials $m_{1},...,m_{s}$. If $s\leq 3$, so that $\Delta$ has at most three minimal nonfaces, or if $\operatorname{supp}(m_{i})\cup\operatorname{supp}(m_{j})=\{x_{1},...,x_{n}\}$ for all $i\neq j$, then $\Delta$ is SCM.

Definition 2.7**.**

Let $I$ be a monomial ideal of $R$. Then the big height of $I$, denoted by $bight(I)$, is $\max\{\operatorname{height}(\mathfrak{p})|\mathfrak{p}\in\operatorname{Ass}(R/I)\}$.

Note that, if $I$ is a matroidal ideal of degree $d$, then by Auslander-Buchasbum formula $bight(I)=n-d+1$.

Proposition 2.8**.**

[21, Corollary 6.4.20]**. Let $I$ be a monomial ideal of $R$ such that $R/I$ is SCM. Then $\operatorname{proj\,dim}(R/I)=bight(I)$.

The following examples say that the converse of Proposition 2.8 is not true even if $I$ is matroidal with $\gcd(I)=1$.

Example 2.9**.**

Let $n=5$ and $I=(x_{1}x_{2},x_{1}x_{3},x_{1}x_{4},x_{1}x_{5},x_{2}x_{3},x_{2}x_{4},x_{3}x_{5},x_{4}x_{5})$ be an ideal of $R$. Then $I$ is a matroidal ideal of $R$ with $\operatorname{proj\,dim}(R/I)=bight(I)$ but $I$ is not SCM.

Proof.

It is clear that $I$ is a matroidal ideal and

[TABLE]

Thus $I^{\vee}_{[3]}=(x_{1}x_{3}x_{4},x_{1}x_{2}x_{5})$ and so $\operatorname{reg}(I^{\vee}_{[3]})=4$. Hence $I^{\vee}$ is not componentwise linear resolution. Therefore $I$ is not SCM but $\operatorname{proj\,dim}(R/I)=4=bight(I)$.

3. SCM matroidal ideals of degree 2

In this section, we classify all SCM matroid ideals of degree 2.

Lemma 3.1**.**

Let $n=3$ and $I$ be a matroidal ideal in $R$ generated in degree $d$. Then $I$ is a SCM ideal.

Proof.

Let $n=3$, then every matroidal ideal in $R$ generated by at most three square-free monomials and so by Corollary 2.6 we have the result.

Lemma 3.2**.**

Let $I$ be a monomial ideal of $R$ such that $I=(u_{1},...,u_{d})$ and $\deg(u_{i})\leq\deg(u_{d})=d$ for all $i$. If $\operatorname{reg}(I)=d$, then $\operatorname{reg}(I_{i})=i$ for all $i>d$.

Proof.

Consider the following exact sequence for $i>d$,

[TABLE]

$l(\dfrac{I}{(I_{i})})<\infty$, so by Lemma 1.1 (d) $\operatorname{reg}(\dfrac{I}{(I_{i})})=i-1$ and

[TABLE]

On the other hand $\operatorname{reg}(I_{i})=\operatorname{reg}(\dfrac{R}{(I_{i})})+1$, that is, $\operatorname{reg}(I_{i})=i$ for all $i>d$.

Proposition 3.3**.**

Let $I$ be monomial ideal which is componentwise linear in $R$. Then $J=(x_{n+1},I)$ is componentwise linear in $R^{\prime}=K[x_{1},...,x_{n},x_{n+1}]$.

Proof.

Suppose that $I=(u_{1},...,u_{m})$, where $\deg(u_{i})=d_{i}$ and $d_{i-1}\leq d_{i}$ for $i=2,...,m$. We induct on $m$, the number of minimal generators of $I$. If $m=1$, then $I=(x_{n+1},u_{1})$. Set $J^{\prime}=x_{n+1}R^{\prime}$. Note that $(J_{j})=(J^{\prime}_{j})$ for all $j<d_{1}$ and so $(J_{j})$ has a linear resolution for all $j<d_{1}$. By Theorem 2.5, $\operatorname{reg}(J)=d_{1}$. Thus $(J_{d_{1}})$ has a linear resolution and also $(J_{j})$ has a linear resolution for all $j>d_{1}$, by using Lemma 3.2.

Now, let $m>1$ and assume that the ideal $L=(x_{n+1},u_{1},...,u_{m-1})$ is componentwise linear. Set $J=(L,u_{m})=(I,x_{n+1})$. Note that $(J_{j})=(L_{j})$ for all $j<d_{m}$ and so $(J_{j})$ has a linear resolution for all $j<d_{m}$. Hence by using [14, Lemma 3.2] we have $\operatorname{reg}(J)=\operatorname{reg}(I)=d_{m}$. Therefore $(J_{d_{m}})$ has a linear resolution. Again, by using Lemma 3.2, we have $(J_{j})$ has a linear resolution for all $j>d_{m}$. This completes the proof.

Corollary 3.4**.**

Let $I$ be a SCM matroidal ideal in $R$ and let $J=x_{n+1}I$ be a monomial ideal in $R^{\prime}=K[x_{1},...,x_{n},x_{n+1}]$. Then $J$ is a SCM matroidal ideal in $R^{\prime}=k[x_{1},...,x_{n},x_{n+1}]$.

Proof.

The Alexander dual of $J$ is $J^{\vee}=(x_{n+1},I^{\vee})$ and by our hypothesis on $I$, $I^{\vee}$ is componentwise linear resolution. Thus by Proposition 3.3, $J^{\vee}$ is componentwise linear resolution. Thus $J$ is a SCM matroidal ideal of $R^{\prime}$.

One of the most distinguished polymatroidal ideals is the ideal of Veronese type. Consider the fixed positive integers $d$ and $1\leq a_{1}\leq...\leq a_{n}\leq d$. The ideal of Veronese type of $R$ indexed by $d$ and $(a_{1},...,a_{n})$ is the ideal $I_{(d;a_{1},...,a_{n})}$ which is generated by those monomials $u=x_{1}^{i_{1}}...x_{n}^{i_{n}}$ of $R$ of degree $d$ with $i_{j}\leq a_{j}$ for each $1\leq j\leq n$.

Remark 3.5**.**

Let $I$ be a SCM matroidal ideal in $R$ and let $J=x_{n+1}...x_{m}I$ be a monomial ideal in $R^{\prime}=K[x_{1},...,x_{n},x_{n+1},...,x_{m}]$. Then, by induction on $m$, $J$ is a SCM matroidal ideal in $R^{\prime}=K[x_{1},...,x_{n},x_{n+1},...,x_{m}]$. Hence for a SCM matroidal ideal $J$, we can assume that $\gcd(J)=1$. By using [16, Lemma 2.16] all fully supported matroid ideals of degree $n-1(n\geq 2)$ are Veronese type ideals and then by Theorem 1.2, all matroidal ideals generated in degrees $d=1,n-1,n$ are SCM.**

Definition 3.6**.**

*Let $I$ be a square-free Veronese ideal of degree $d$. We say that $J$ is an almost square-free Veronese ideal of degree $d$ when $J\neq 0$, $G(J)\subseteq G(I)$ and

${\mid G(J)\mid\geq\mid G(I)\mid-1}$. Note that every square-free Veronese ideal is an almost quare-free Veronese ideal. Also, if $J$ is an almost square-free Veronese ideal of degree $n$, then $J$ is a square-free Veronese ideal.*

Lemma 3.7**.**

Let $J$ be an almost square-free Veronese ideal of degree $d<n$. Then $J$ is a SCM matroidal ideal of $R$.

Proof.

Suppose that $y_{1},...,y_{n}$ is an arbitrary permutation of the variables of $R$ such that $\{y_{1},...,y_{n}\}=\{x_{1},...,x_{n}\}$ and let $I$ be a square-free Veronese ideal of degree $d$. We may assume that $I=J+(y_{n-d+1}y_{n-d+2}...y_{n})$. Then we have $J=(y_{1},...,y_{n-d})\cap I$ and so $J$ is a matroidal ideal. Therefore $J^{\vee}=(y_{1}...y_{n-d},I^{\vee})$. Set $J^{\prime}=(y_{1}...y_{n-d})$. Then, for all $i\leq n-d$, ${J^{\vee}}_{[i]}=J^{\prime}_{[i]}$ and so it is componentwise linear. For all $i\geq n-d+1$, ${J^{\vee}}_{[i]}$ is a square-free Veronese ideal and so $J^{\vee}$ is a componentwise linear ideal. Hence $J$ is a SCM matroidal ideal, as required.

From now on, we will let $y_{1},...,y_{n}$ be an arbitrary permutation variables of $R$ such that $\{x_{1},...,x_{n}\}=\{y_{1},...,y_{n}\}$.

Theorem 3.8**.**

Let $J$ be a matroidal ideal of $R$ with $\deg(J)=2$ and $\gcd(J)=1$. Then $J$ is SCM if and only if there exists a permutation of variables such that the following hold:

• (a)

$J=y_{1}\mathfrak{p}+J^{\prime}$, where $\mathfrak{p}$ is a monomial prime ideal with $y_{1}\notin\mathfrak{p}$, $\operatorname{height}(\mathfrak{p})=n-1$ and $J^{\prime}$ is a SCM matroidal ideal with $\operatorname{supp}(J^{\prime})=\{y_{2},...,y_{n}\}$ and $\gcd(J^{\prime})=1$, or

• (b)

$J=y_{1}\mathfrak{p}+y_{2}\mathfrak{q}$, where $\mathfrak{p}$ and $\mathfrak{q}$ are monomial prime ideals with $y_{1}\notin\mathfrak{p}$ and $y_{1},y_{2}\notin\mathfrak{q}$ such that $\operatorname{height}(\mathfrak{p})=n-1$, $\operatorname{height}(\mathfrak{q})=n-2$.

Proof.

$(\Longleftarrow).$ Consider the case $(a)$. We have $J=\mathfrak{p}\cap(y_{1},J^{\prime})$, then $J^{\vee}=(\mathfrak{p}^{\vee},y_{1}J^{\prime\vee})$ and $\mathfrak{p}^{\vee}\in(u)$ for all $u\in J^{\prime\vee}$. Since ${J^{\vee}}_{[i]}=y_{1}{J^{\prime\vee}}_{[i-1]}$ for all $i\leq n-2$, and ${J^{\prime\vee}}_{[i-1]}$ is componentwise linear, it follows that ${J^{\vee}}_{[i]}$ is componentwise linear for all $i\leq n-2$. now consider the exact sequence

[TABLE]

From $(y_{1}J^{\prime\vee}:\mathfrak{p}^{\vee})=(y_{1})$, we have $\operatorname{reg}(R/(y_{1}J^{\prime\vee}:\mathfrak{p}^{\vee}))=0$. Since $\deg(\mathfrak{p}^{\vee})=n-1$, we have $\operatorname{reg}(R/(\mathfrak{p}^{\vee},y_{1}J^{\prime\vee}))\geq n-2$. Since $y_{1}J^{\prime\vee}$ is componentwise linear and $\deg(u)\leq n-2$ for all $u\in J^{\prime\vee}$, by [11, Corollary 8.2.14] we have $\operatorname{reg}(R/y_{1}J^{\prime\vee})\leq n-2$. By using Lemma 1.1,

[TABLE]

It therefore follows $\operatorname{reg}(R/(\mathfrak{p}^{\vee},y_{1}J^{\prime\vee}))=n-2$. Thus ${J^{\vee}}_{[n-1]}$ has a linear resolution and so $J$ is a SCM ideal.

Let us consider the case $(b)$. $J=(y_{1},y_{2})\cap(y_{1},\mathfrak{q})\cap\mathfrak{p}$ and so $J^{\vee}=(y_{1}y_{2},y_{1}\mathfrak{q}^{\vee},\mathfrak{p}^{\vee})$. It is clear that $J^{\vee}$ is a monomial ideal with linear quotients. Thus, by Proposition 2.2, $J^{\vee}$ is componentwise linear and so $J$ is a SCM ideal.

$(\Longrightarrow).$ Let $J$ be a SCM ideal. Then there exists $\mathfrak{p}\in\operatorname{Ass}(R/J)$ such that $\operatorname{height}(\mathfrak{p})=\operatorname{proj\,dim}(R/J)=n-1$. Since $J=\cap_{i=1}^{n}(J:y_{i})$ and $\deg(J)=2$, we can consider $\mathfrak{p}=(J:y_{1})$ and $\mathfrak{p}=(y_{2},...,y_{n})$. Hence $J=y_{1}\mathfrak{p}+J^{\prime}$, where $J^{\prime}$ is a matroidal ideal of degree $2$ in $K[y_{2},...,y_{n}]$. We claim that $\operatorname{supp}(J^{\prime})=\{y_{2},...,y_{n}\}$. Let $y_{l}\notin\operatorname{supp}(J^{\prime})$, where $l\geq 2$. Thus $y_{1}y_{l},y_{j}y_{k}\in J$, where $j,k\geq 2$. Since $J$ is a matroidal ideal, it follows $y_{l}y_{k}$ or $y_{l}y_{j}\in J$. Hence $y_{l}y_{k}$ or $y_{l}y_{j}\in J^{\prime}$ and this is a contradiction. Therefore $\operatorname{supp}(J^{\prime})=\{y_{2},...,y_{n}\}$. $J=\mathfrak{p}\cap(J^{\prime},y_{1})$, it follows that $J^{\vee}=(\mathfrak{p}^{\vee},y_{1}J^{\prime\vee})$. For all $i\leq n-2$, we have ${J^{\vee}}_{[i]}=y_{1}{J^{\prime\vee}}_{[i-1]}$ and so ${J^{\prime\vee}}_{[i-1]}$ has a linear resolution for all $i\leq n-2$. Since ${J^{\vee}}_{[n-1]}=y_{1}{J^{\prime\vee}}_{[n-2]}+(\mathfrak{p}^{\vee})$ and $\operatorname{reg}({J^{\vee}}_{[n-1]})=n-1$, it follows that $\operatorname{reg}({J^{\prime\vee}}_{[n-2]})\leq n-2$. Therefore ${J^{\prime\vee}}_{[n-2]}$ has a linear resolution and so $J^{\prime\vee}$ is componentwise linear. That is $J^{\prime}$ is a SCM matroidal ideal of degree $2$. If $\gcd(J^{\prime})=1$, then $J$ satisfy in the case $(a)$. If $\gcd(J^{\prime})\neq 1$, then we have the case $(b)$. This completes the proof.

4. SCM matroidal ideals over polynomial rings of small dimensional

We start this section by the following fundamental lemma.

Lemma 4.1**.**

Let $n\geq 5$ and $J$ be a matroidal ideal of degree $d$ in $R$ and $\gcd(J)=1$. If $J$ is SCM, then

[TABLE]

where $\mathfrak{p}=(y_{d},...,y_{n})$ is a monomial prime ideal, $J_{i}$ is a SCM matroidal ideal of degree $2$ with $\operatorname{supp}(J_{i})=\{y_{d},y_{d+1},...,y_{n}\}$ for $i=1,...,d-1$ and $J_{d}\subseteq\cap_{i=1}^{d-1}J_{i}$.

Proof.

$J$ is a SCM matroidal ideal, then there is a prime ideal $\mathfrak{p}\in\operatorname{Ass}(R/J)$ such that $\operatorname{height}(\mathfrak{p})=\operatorname{proj\,dim}(R/J)$. Since $\operatorname{depth}(R/J)=d-1$, it follows that $\operatorname{height}(\mathfrak{p})=n-d+1$. For every square-free monomial ideal in $R$, we have $J=\cap_{i=1}^{n}(J:y_{i})$. It follows that $\mathfrak{p}=(J:y_{1}y_{2}...y_{d-1})$ and we can write $J=y_{1}...y_{d-1}\mathfrak{p}+J^{\prime}$, where $J^{\prime}$ is a square-free monomial ideal of degree $d$. It is clear that $J^{\prime}$ has a presentation

[TABLE]

and $J_{d}\subseteq\cap_{i=1}^{d-1}J_{i}$. Note that $\gcd(J)=1$ and

[TABLE]

we have $\operatorname{height}(J)\geq 2$ and so $J_{i}\neq 0$ for $i=1,...,d-1$. It is known that the localization of every SCM ideal is SCM and so

[TABLE]

is a SCM matroidal ideal of degree $2$ for $i=1,...,d-1$. By using the proof of Theorem 3.8, $J_{i}$ is a SCM matroidal ideal with $\operatorname{supp}(J_{i})=\{y_{d},y_{d+1},...,y_{n}\}$ for $i=1,...,d-1$.

It is known that the localization of each SCM matroidal ideal is a SCM matroidal ideal. The following example shows that the converse is not true.

Example 4.2**.**

Let $n=4$ and $J=(x_{1}x_{3},x_{1}x_{4},x_{2}x_{3},x_{2}x_{4})$. Then $J$ is a matroidal ideal and $(J:x_{i})$ is SCM matroidal for $i=1,2,3,4$; but $J$ is not SCM.

Proof.

It is clear that $J$ is matroidal and $(J:x_{i})$ is SCM matroidal for $i=1,2,3,4$. Since $J^{\vee}=(x_{1}x_{2},x_{3}x_{4})$, it follows that $\operatorname{reg}(J^{\vee})=3$. Therefore $J$ is not SCM.

From now on, as Lemma 4.1, for a SCM matroidal ideal $J$ of degree $d$ and $\gcd(J)=1$ in $R$ with $n\geq 5$, we can write

[TABLE]

where $\mathfrak{p}=(y_{d},...,y_{n})$ is a monomial prime ideal, $J_{i}$ is a SCM matroidal ideal of degree $2$ with $\operatorname{supp}(J_{i})=\{y_{d},y_{d+1},...,y_{n}\}$ for $i=1,...,d-1$ and $J_{d}\subseteq\cap_{i=1}^{d-1}J_{i}$.

Note that if for instance $\gcd(J_{1})=y_{d}$, then we have

[TABLE]

where $\mathfrak{q}=(y_{d+1},...,y_{n})$.

Bandari and Herzog in [1, Proposition 2.7] proved that if $n=3$ and $J$ is a matroidal ideal with $\gcd(J)=1$, then $J$ is a square-free Veronese ideal and so by Theorem 1.2, it is CM (see also [17, Proposition 1.5]). In the following proposition we prove this result in the case $n=4$ for SCM ideals.

Proposition 4.3**.**

Let $n=4$ and $J$ be a matroidal ideal of $R$ of degree $d$ and $\gcd(J)=1$. Then $J$ is a SCM ideal if and only if $J$ is

• (a)

a square-free Veronese ideal, or

• (b)

an almost square-free Veronese ideal.

Proof.

$(\Longleftarrow)$ is clear by Theorem 1.2 and Lemma 3.7.

$(\Longrightarrow)$. If $d=1,3,4$, then by Theorem 1.2 and [16, Lemma 2.16] $J$ is a square-free Veronese ideal. If $d=2$, then by Theorem 3.8, $J=y_{1}\mathfrak{p}+J^{\prime}$, where $\mathfrak{p}$ is a monomial prime ideal with $y_{1}\notin\mathfrak{p}$, $\operatorname{height}(\mathfrak{p})=3$ and $J^{\prime}$ is a SCM matroidal ideal with $\operatorname{supp}(J^{\prime})=\{y_{2},y_{3},y_{4}\}$. If $\gcd(J^{\prime})=1$, then $J^{\prime}$ is a square-free Veronese ideal and so is $J$. If $\gcd(J^{\prime})\neq 1$, then $J^{\prime}$ is an almost square-free Veronese ideal.

Proposition 4.4**.**

Let $n=4$ and $J$ be a matroidal ideal of $R$ of degree $d$. Then $J$ is a SCM ideal if and only if $\operatorname{proj\,dim}(R/J)=bight(J)$.

Proof.

$(\Longrightarrow)$. It follows by Proposition 2.8.

$(\Longleftarrow)$. If $d=1,3,4$, then by Remark 3.5 $J$ is SCM. Let $d=2$. By our hypothesis, there exists $\mathfrak{p}\in\operatorname{Ass}(R/J)$ such that $\mathfrak{p}=(J:y_{1})$. Thus $J=y_{1}\mathfrak{p}+J^{\prime}$, where $J^{\prime}$ is matroidal ideal of degree $2$ in $K[y_{2},y_{3},y_{4}]$. Hence $J^{\prime}$ is a square-free Veronese ideal or an almost square-free Veronese ideal. Therefore by Proposition 4.3, $J$ is SCM.

Lemma 4.5**.**

Let $n\geq 5$ and $J$ be a matroidal ideal of degree $3$ in $R$ such that $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}y_{3}\mathfrak{q}$, where $\mathfrak{p}$ and $\mathfrak{q}$ are monomial prime ideals with $y_{1},y_{2}\notin\mathfrak{p}$ and $y_{1},y_{2},y_{3}\notin\mathfrak{q}$ such that $\operatorname{height}(\mathfrak{p})=n-2$, $\operatorname{height}(\mathfrak{q})=n-3$. Then $J$ is SCM.

Proof.

Since $J=\mathfrak{p}\cap(y_{1}y_{2},y_{1}y_{3}\mathfrak{q},y_{2}y_{3}\mathfrak{q})$, it follows that $J=\mathfrak{p}\cap(y_{1},y_{2})\cap(y_{1},y_{3})\cap(y_{2},y_{3})\cap(y_{1},\mathfrak{q})\cap(y_{2},\mathfrak{q})$. Therefore $J^{\vee}=(y_{1}y_{2},y_{1}y_{3},y_{2}y_{3},y_{1}{\mathfrak{q}}^{\vee},y_{2}{\mathfrak{q}}^{\vee},\mathfrak{p}^{\vee})$. It is clear that $J^{\vee}$ is a monomial ideal with linear quotients and so by Proposition 2.2, $J^{\vee}$ is componentwise linear. Thus $J$ is SCM.

Lemma 4.6**.**

Let $n\geq 5$ and $J$ be a matroidal ideal of degree $3$ such that

[TABLE]

where $\mathfrak{p}$, $\mathfrak{q}_{1}$ and $\mathfrak{q}_{2}$ are monomial prime ideals with $y_{1},y_{2}\notin\mathfrak{p}$, $y_{1},y_{2},y_{3}\notin\mathfrak{q}_{1}$ and $y_{1},y_{2},y_{4}\notin\mathfrak{q}_{2}$ such that $\operatorname{height}(\mathfrak{p})=n-2$, $\operatorname{height}(\mathfrak{q}_{1})=n-3=\operatorname{height}(\mathfrak{q}_{2})$ and $J_{1}$ is a matroidal ideal in $R^{\prime}=K[y_{3},...,y_{n}]$. Then $G(J_{1})=\{y_{3}y_{4}y_{i}\mid i=5,6,...,n\}$. In particular, $J$ is not SCM.

Proof.

We consider two cases:

** Case (a): **

$J_{1}=0$, then we have $y_{1}y_{3}y_{5},y_{2}y_{3}y_{4}\in J$ but $y_{2}y_{3}y_{5}$ or $y_{3}y_{4}y_{5}$ are not elements of $J$. Thus $J$ is not a matroidal ideal and this is a contradiction.

** Case (b): **

$J_{1}\neq 0$.

**1): **

For $n=5$, $J_{1}=(y_{3}y_{4}y_{5})$ and

[TABLE]

Therefore $\operatorname{reg}(J^{\vee}_{[2]})=3$ and so $J$ is not SCM.

**2): **

Suppose that $n\geq 6$. Then $(J:y_{3})=(y_{1}y_{2},y_{2}y_{4},y_{1}\mathfrak{q}_{1},(J_{1}:y_{3}))$. If $y_{i}y_{j}\in(J:y_{3})$ for $5\leq i\neq j\leq n$, then $y_{2}y_{i}\in(J:y_{3})$ for $i\geq 5$, since $y_{2}y_{4}\in(J:y_{3})$. But this is a contradiction. Therefore $y_{3}y_{i}y_{j}\notin J$ for all $5\leq i\neq j\leq n$. Consider $(J:y_{4})$, we have $y_{4}y_{i}y_{j}\notin J$ for all $5\leq i\neq j\leq n$. Also, if $y_{i}y_{j}y_{t}\in J$ for different numbers $i,j,t$ with $5\leq i,j,t\leq n$, then since $y_{1}y_{3}y_{i}\in J$, we have $y_{3}y_{i}y_{j}\in J$ or $y_{3}y_{i}y_{t}\in J$ and this is a contradiction. Thus $G(J_{1})\subseteq\{y_{3}y_{4}y_{i}\mid i=5,6,...,n\}$. On the other hand, since $y_{2}y_{4}y_{i}$ and $y_{1}y_{3}y_{i}$ are elements in $J$ for $i\geq 5$ we have $y_{3}y_{4}y_{i}\in J$ for $i\geq 5$. Hence $G(J_{1})=\{y_{3}y_{4}y_{i}\mid i=5,6,...,n\}$. Therefore

[TABLE]

and so $J^{\vee}=(y_{1}y_{4},y_{2}y_{3},y_{1}y_{2}J_{1}^{\vee},y_{1}\mathfrak{q}_{2}^{\vee},y_{2}\mathfrak{q}_{1}^{\vee},\mathfrak{p}^{\vee})$. Thus $\operatorname{reg}(J^{\vee}_{[2]})=3$ and so $J$ is not SCM.

Lemma 4.7**.**

Let $n\geq 6$ and $J$ be a matroidal ideal of degree $3$ such that $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}J_{1}$, where $\mathfrak{p}$ and $\mathfrak{q}$ are monomial prime ideals with $y_{1},y_{2}\notin\mathfrak{p}$, $y_{1},y_{2},y_{3}\notin\mathfrak{q}$ such that $\operatorname{height}(\mathfrak{p})=n-2$, $\operatorname{height}(\mathfrak{q})=n-3$ and $J_{1}$ is a matroidal ideal in $R^{\prime}=K[y_{3},...,y_{n}]$ with $\gcd(J_{1})=1$. Then $J$ is not SCM matroidal.

Proof.

By contrary, we assume that $J$ is SCM matroidal. Then $(J:y_{2})=y_{1}\mathfrak{p}+J_{1}$ is SCM matroidal and so by Theorem 3.8 $J_{1}$ is SCM matroidal of degree $2$. From $\gcd(J_{1})=1$, we have $J_{1}=y_{i}\mathfrak{q}_{1}+J_{2}$, where $\mathfrak{q}_{1}$ and $J_{2}$ are a monomial prime ideal of height $n-3$ and a matroidal ideal respectively in $R^{\prime}=K[y_{3},...,y_{i-1},y_{i+1},...,y_{n}]$. There are two main cases to consider.

**a): **

$i=3$, then $(J:y_{j})=(y_{1}y_{2},y_{1}y_{3},y_{2}y_{3},y_{2}(J_{2}:y_{j}))$ when $j\neq 1,2,3$. Since $y_{t}\in(J_{2}:y_{j})$ for $t\neq 1,2,3,j$, we have $y_{2}y_{t}$ and $y_{1}y_{3}$ are elements of $(J:y_{j})$ but $y_{1}y_{t}$ or $y_{3}y_{t}$ are not elements of $(J:y_{j})$. This is a contradiction.

**b): **

$i\neq 3$, then $(J:y_{i})=(y_{1}y_{2},y_{1}y_{3},y_{2}\mathfrak{q}_{1})$. Thus $y_{2}y_{t}$ and $y_{1}y_{3}$ for $t\neq 3$ are elements of $(J:y_{i})$ but $y_{1}y_{t}$ or $y_{3}y_{t}$ are not elements of $(J:y_{i})$ and this is a contradiction. Thus $J$ is not SCM matroidal.

Lemma 4.8**.**

Let $n\geq 6$ and $J$ be a matroidal ideal of degree $3$ such that $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}J_{1}+J_{2}$ or $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}y_{3}\mathfrak{q}+J_{2}$, where $\mathfrak{p}$ and $\mathfrak{q}$ are monomial prime ideals with $y_{1},y_{2}\notin\mathfrak{p}$, $y_{1},y_{2},y_{3}\notin\mathfrak{q}$ such that $\operatorname{height}(\mathfrak{p})=n-2$, $\operatorname{height}(\mathfrak{q})=n-3$ and $J_{1}$ is a nonzero matroidal ideal in $R^{\prime}=K[y_{3},...,y_{n}]$ with $\gcd(J_{1})=1$. Then $G(J_{2})\subseteq\{y_{3}y_{i}y_{j}\mid 4\leq i\neq j\leq n\}$ and if $J_{2}\neq 0$, then $\operatorname{supp}(J_{2})=\{y_{3},y_{4},...,y_{n}\}$. In particular, if $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}y_{3}\mathfrak{q}+J_{2}$, then $J_{2}=0$.

Proof.

Let us consider $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}J_{1}+J_{2}$. Then we have $(J:y_{t})=(y_{1}y_{2},y_{1}y_{3},y_{2}(J_{1}:y_{t}),(J_{2}:y_{t}))$ for some $t\geq 4$. If $y_{i}y_{j}y_{t}\in J$ for some different numbers $4\leq i,j,t\leq n$, then $y_{i}y_{j}\in(J:y_{t})$. Since $y_{1}y_{3}\in(J:y_{t})$, it follows that $y_{1}y_{i}\in(J:y_{t})$ for some $i\geq 4$ and this is a contradiction. It therefore follows that $G(J_{2})\subseteq\{y_{3}y_{i}y_{j}\mid 4\leq i\neq j\leq n\}$. Also, $(J:y_{3})=(y_{1}y_{2},y_{1}\mathfrak{q},y_{2}(J_{1}:y_{3}),(J_{2}:y_{3}))$. If $y_{i}y_{j}\in(J:y_{3})$ for some $4\leq i\neq j\leq n$, then $y_{i}y_{t}\in(J:y_{3})$ for all $t$ with $4\leq i\neq t\leq n$ since $y_{1}y_{t}\in(J:y_{3})$. Hence $\operatorname{supp}(J_{2})=\{y_{3},y_{4},...,y_{n}\}$. The proof for the case $J=y_{1}y_{2}\mathfrak{p}+y_{1}y_{3}\mathfrak{q}+y_{2}y_{3}\mathfrak{q}+J_{2}$ is similar to the above argument. In particular, if $y_{3}y_{i}y_{j}\in J_{2}$ for some $4\leq i\neq j\leq n$ then from $y_{1}y_{2}y_{t}\in J$ for some $4\leq i\neq t\neq j\leq n$ we have $y_{i}y_{j}y_{t}\in J$. This is a contradiction. Thus $J_{2}=0$.

Proposition 4.9**.**

Let $n=5$ and $J$ be a matroidal ideal of degree $3$ such that $\gcd(J)=1$. Then $J$ is a SCM ideal if and only if $J=y_{1}y_{2}\mathfrak{p}+y_{1}J_{1}+y_{2}J_{2}+J_{3}$, where $J_{1}$ and $J_{2}$ are SCM ideals with $\operatorname{supp}(J_{1})=\operatorname{supp}(J_{2})=\{y_{3},y_{4},y_{5}\}$, $J_{3}\subseteq J_{1}\cap J_{2}$ and satisfying in the one of the following cases:

• (a)

$\gcd(J_{1})=1$, $\gcd(J_{2})=1$, or

• (b)

$\gcd(J_{1})=y_{3}=\gcd(J_{2})$* and $J_{3}=0$.*

Proof.

$(\Longleftarrow)$. Consider $(a)$. Then $J_{1}$ and $J_{2}$ are square-free Veronese ideal and $G(J_{3})\subseteq\{y_{3}y_{4}y_{5}\}$. If $J_{3}=0$, then $J$ is an almost square-free Veronese ideal and so by using Lemma 3.7, $J$ is a SCM matroidal ideal. If $J_{3}\neq 0$, then $J$ is a square-free Veronese ideal and so $J$ is a SCM matroidal ideal.

If we have the case $(b)$, then by Lemma 4.5 the result follows.

$(\Longrightarrow).$ Let $J$ be a SCM, then by Lemma 4.1, $J$ has the presentation $J=y_{1}y_{2}\mathfrak{p}+y_{1}J_{1}+y_{2}J_{2}+J_{3}$, where $J_{1}$ and $J_{2}$ are SCM matroidal ideals with $\operatorname{supp}(J_{1})=\operatorname{supp}(J_{2})=\{y_{3},y_{4},y_{5}\}$ and $J_{3}\subseteq J_{1}\cap J_{2}$.

**1): **

If $\gcd(J_{1})=y_{3}$ and $\gcd(J_{2})=y_{4}$, then by Lemma 4.6 $J$ is not a SCM matroidal ideal and we don’t have this case.

**2): **

If $\gcd(J_{1})=\gcd(J_{2})=y_{3}$, then $J_{3}=0$. Let contrary, then $G(J_{3})=\{y_{3}y_{4}y_{5}\}$ and $y_{1}y_{2}y_{5},y_{3}y_{4}y_{5}\in J$ but $y_{1}y_{4}y_{5}$ or $y_{2}y_{4}y_{5}$ are not elements of $J$. This is a contradiction.

**3): **

If $\gcd(J_{1})=y_{3}$, $\gcd(J_{2})=1$ and $J_{3}=0$, then $y_{1}y_{3}y_{5},y_{2}y_{4}y_{5}\in J$ but $y_{1}y_{4}y_{5}$ or $y_{3}y_{4}y_{5}$ are not elements of $J$. Therefore $J$ is not matroidal and we don’t have this case.

**4): **

If $\gcd(J_{1})=y_{3}$, $\gcd(J_{2})=1$ and $G(J_{3})=\{y_{3}y_{4}y_{5}\}$, then by change of variables $(a)$ follows with $J_{3}=0$.

Proposition 4.10**.**

Let $n=6$ and let $J$ be a matroidal ideal of degree $4$ such that $\gcd(J)=1$. Then $J$ is a SCM ideal if and only if $J=y_{1}y_{2}y_{3}\mathfrak{p}+y_{1}y_{2}J_{1}+y_{1}y_{3}J_{2}+y_{2}y_{3}J_{3}+J_{4}$ such that $J_{1},J_{2},J_{3}$ are SCM matroidal ideals and satisfying in one of the following conditions:

• (a)

for $i=1,2,3$, $\gcd(J_{i})=1$ and $\mid G(J_{4})\mid=3$,

• (b)

for $i=1,2,3$, $\gcd(J_{i})=1$ and $\mid G(J_{4})\mid=2$,

• (c)

for $i=1,2,3$, $\gcd(J_{i})=1$ and $J_{4}=0$, or

• (d)

for $i=1,2,3$, $\gcd(J_{i})=y_{4}$ and $J_{4}=0$.

Proof.

$(\Longleftarrow).$ If we have $(a)$, then $J$ is a square-free Veronese ideal and so by Theorem 1.2, $J$ is SCM. Consider case $(b)$, then $J$ is an almost square-free Veronese ideal and so by Lemma 3.7, $J$ is SCM. If we consider $(d)$, then by using the same proof of Lemma 4.5 $J^{\vee}$ has linear quotients and so $J$ is SCM. Let $(c)$, then we have $J=\mathfrak{p}\cap(y_{1},y_{2})\cap(y_{1},y_{3})\cap(y_{2},y_{3})\cap(y_{1},J_{3})\cap(y_{2},J_{2})\cap(y_{3},J_{1})$ and so $J^{\vee}=(y_{1}y_{2},y_{1}y_{3},y_{2}y_{3},y_{1}J^{\vee}_{3},y_{2}J^{\vee}_{2},y_{3}J^{\vee}_{1},\mathfrak{p}^{\vee})$. That is, $J^{\vee}$ has linear quotients. Thus $J$ is SCM.

$(\Longrightarrow).$ Let $J$ be a SCM ideal. Then by Lemma 4.1, $J=y_{1}y_{2}y_{3}\mathfrak{p}+y_{1}y_{2}J_{1}+y_{1}y_{3}J_{2}+y_{2}y_{3}J_{3}+J_{4}$ and $J_{1},J_{2},J_{3}$ are SCM matroidal ideals. Let $\gcd(J_{1})=y_{4}$. Since $(J:y_{1})=y_{2}y_{3}\mathfrak{p}+y_{2}J_{1}+y_{3}J_{2}+(J_{4}:y_{1})$, $\gcd(J:y_{1})=1$ and $(J:y_{1})$ is a SCM matroidal ideal, by Proposition 4.9 it follows $\gcd(J_{2})=y_{4}$ and $(J_{4}:y_{1})=0$. Again by using $(J:y_{2})$ and $(J:y_{3})$, we obtain $\gcd(J_{1})=\gcd(J_{3})=\gcd(J_{2})=y_{4}$ and $J_{4}=0$. Also, if for some $i$, $\gcd(J_{i})=1$, then by Proposition 4.9 and by using $(J:y_{1}),(J:y_{2})$ and $(J:y_{3})$ we have $\gcd(J_{i})=1$ for $i=1,2,3$. If $G(J_{4})=\{y_{1}y_{4}y_{5}y_{6}\}$, then $J$ is not a matroidal ideal since $y_{1}y_{4}y_{5}y_{6},y_{2}y_{3}y_{5}y_{6}\in J$, but $y_{2}y_{4}y_{5}y_{6}$ or $y_{3}y_{4}y_{5}y_{6}$ are not elements of $J$. Thus $J_{4}=0$ or $\mid G(J_{4})\mid=2$ or $\mid G(J_{4})\mid=3$ and this completes the proof.

Proposition 4.11**.**

Let $n\geq 6$ and let $J$ be a matroidal ideal of degree $n-2$ such that $\gcd(J)=1$. Then $J$ is a SCM ideal if and only if

[TABLE]

such that $J_{i}$ is SCM matroidal ideal for all $i=1,..,n-3$ and satisfying in one of the following conditions:

• (a)

for $i=1,...,n-3$, $\gcd(J_{i})=1$ and $\mid G(J_{n-2})\mid=\binom{n-3}{2}$,

• (b)

for $i=1,...,n-3$, $\gcd(J_{i})=1$ and $\mid G(J_{n-2})\mid=\binom{n-3}{2}-1$,

• (c)

for $i=1,...,n-3$, $\gcd(J_{i})=1$ and $J_{n-2}=0$, or

• (d)

for $i=1,...n-3$, $\gcd(J_{i})=y_{n-2}$ and $J_{n-2}=0$.

Proof.

$(\Longleftarrow).$

If case $(a)$ holds, then $J$ is a square-free Veronese ideal and so by Theorem 1.2, $J$ is SCM. Let $(b)$, then $J$ is an almost square-free Veronese ideal and so by Lemma 3.7, $J$ is SCM. If $(d)$, then by using the same proof of Lemma 4.5, $J^{\vee}$ has linear quotients and so $J$ is SCM. Let $(c)$, then we have

[TABLE]

and so

[TABLE]

Since $J_{i}$ are square-free Veronese ideals, it follows that $J^{\vee}$ has linear quotients. That is, $J$ is SCM.

$(\Longrightarrow).$ Let $J$ be a SCM ideal. Then by Lemma 4.1,

[TABLE]

and $J_{i}$ are SCM matroidal ideals for all $i=1,...,n-3$. We use induction on $n\geq 6$. If $n=6$, then the result follows by Proposition 4.10. Let $n>6$ and $\gcd(J_{1})=y_{n-2}$.

[TABLE]

$\gcd(J:y_{1})=1$ and $(J:y_{1})$ is a SCM matroidal ideal, by induction hypothesis it follows $\gcd(J_{i})=y_{n-2}$ for $i=1,...,n-4$ and $(J_{n-2}:y_{1})=0$. Again by using $(J:y_{i})$ for $i=2,...,n-3$ and by using induction hypothesis, $\gcd(J_{i})=y_{n-2}$ for $i=1,...n-3$ and $J_{n-2}=0$. Also, if for some $i$, $\gcd(J_{i})=1$, then again by using $(J:y_{i})$ for $i=1,...,n-3$ and by using induction hypothesis we have $\gcd(J_{i})=1$ for $i=1,...,n-3$. If $\mid G(J_{n-2})\mid<\binom{n-3}{2}-1$, then there exists $1\leq i\leq{n-3}$ such that $\mid G(I:y_{i})\mid<\binom{n-4}{2}-1$ and this is a contradiction. Thus $J_{n-2}=0$ or $\mid G(J_{n-2})\mid=\binom{n-3}{2}$ or $\mid G(J_{n-2})\mid=\binom{n-3}{2}-1$ and this completes the proof.

Theorem 4.12**.**

Let $n=6$ and let $J$ be a matroidal ideal of degree $3$ such that $\gcd(J)=1$. Then $J$ is a SCM ideal if and only if $J=y_{1}y_{2}\mathfrak{p}+y_{1}J_{1}+y_{2}J_{2}+J_{3}$ such that $J_{1}$ and $J_{2}$ are SCM matroidal ideals and satisfying in one of the following conditions:

• (a)

$\mid G(J_{3})\mid=4$* and one of $J_{1}$ or $J_{2}$ is an almost square-free Veronese ideal and the other is a square-free Veronese ideal,*

• (b)

$\mid G(J_{3})\mid=3$, $J_{1}$, $J_{2}$ are square-free Veronese ideals,

• (c)

$J_{3}=0$* and $J_{1}=J_{2}$ are square-free Veronese ideals or almost square-free Veronese ideals either $J_{3}=0$ and $\gcd(J_{1})=y_{3}=\gcd(J_{2})$.*

Proof.

$(\Longleftarrow)$. If we consider the $(a)$ or $(b)$, then $J$ is a square-free Veronese ideal or an almost square-free Veronese ideal and so $J$ is SCM. Consider $(c)$ and suppose that $\gcd(J_{1})=\gcd(J_{2})=y_{3}$. Then by using Lemma 4.5, $J$ is SCM. Also, for $(c)$ if $J_{1}=J_{2}$ are square-free Veronese ideals or almost square-free Veronese ideals, we have $J^{\vee}=(y_{1}y_{2},y_{1}J^{\vee}_{2},y_{2}J^{\vee}_{1},\mathfrak{p}^{\vee})$ and so $J^{\vee}$ has linear quotients. Thus $J$ is SCM.

$(\Longrightarrow).$ Let $J$ be a SCM ideal. Then by Lemma 4.1, $J=y_{1}y_{2}\mathfrak{p}+y_{1}J_{1}+y_{2}J_{2}+J_{3}$ and $J_{1}$ and $J_{2}$ are SCM matroidal ideals and $J_{3}\subseteq J_{1}\cap J_{2}$ with $\suppressfloats(J_{3})=\{y_{3},y_{4},y_{5},y_{6}\}$. Therefore $\mid G(J_{3})\mid\leq 4$. We have four cases:

**Case (i): **

Suppose that $\mid G(J_{3})\mid=4$ ,then by Lemmas 4.6 and 4.8 we have $\gcd(J_{1})=1=\gcd(J_{2})$. By Proposition 4.3, we have the case $(a)$ if we prove $J_{1}$ and $J_{2}$ aren’t almost square-free Veronese ideals in the same time. Let contrary, if $y_{1}y_{3}y_{5},y_{2}y_{3}y_{5}$ are not elements of $J$, then $y_{1}y_{2}y_{3},y_{3}y_{4}y_{5}\in J$. But $y_{1}y_{3}y_{5}$ or $y_{2}y_{3}y_{5}$ are not elements of $J$ and this is a contradiction.

If $y_{1}y_{3}y_{5},y_{2}y_{3}y_{6}$ are not elements of $J$, then

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By Theorem 3.8, $(y_{1}y_{2},y_{1}y_{6},y_{2}y_{5},y_{5}y_{6})$ is not SCM and this is a contradiction.

If $y_{1}y_{3}y_{5},y_{2}y_{4}y_{6}$ are not elements of $J$, then $(J^{\vee}_{[3]})=(y_{1}y_{3}y_{5},y_{2}y_{4}y_{6})$ and so $\operatorname{reg}(J^{\vee}_{[3]})=5$. Thus $J$ is not SCM and this is a contradiction.

**Case (ii): **

Let $\mid G(J_{3})\mid=3$. We consider the following cases.

**1): **

If $\gcd(J_{1})=y_{3}$ and $\gcd(J_{2})=1$, then $G(J_{3})=\{y_{3}y_{4}y_{5},y_{3}y_{4}y_{6},y_{3}y_{5}y_{6}\}$, by Lemma 4.8. $\gcd(J_{2})=1$, so by Proposition 4.3, $J_{2}$ is a square-free Veronese ideal or an almost square-free Veronese ideal. If $J_{2}=(y_{2}y_{3}y_{4},y_{2}y_{3}y_{5},y_{2}y_{3}y_{6},y_{2}y_{4}y_{5},y_{2}y_{4}y_{6})$ is an almost square-free Veronese ideal, then $y_{3}y_{5}y_{6},y_{1}y_{2}y_{4}\in J$ but $y_{1}y_{5}y_{6}$ or $y_{2}y_{5}y_{6}$ either $y_{4}y_{5}y_{6}$ are not elements of $J$ and this is a contradiction. So $J_{2}$ is a square-free Veronese ideal and by using a new presentation for $J$ and change of variables we get $J_{1}$ and $J_{2}$ are square-free Veronese ideals and $J_{3}=0$ and this is the case $(c)$.

**2): **

If $\gcd(J_{1})=y_{3}$ and $\gcd(J_{2})=y_{4}$, then by Lemma 4.6 we have $\mid G(J_{3})\mid=2$ and this is a contradiction.

**3): **

If $\gcd(J_{1})=y_{3}=\gcd(J_{2})$, then $y_{1}y_{2}y_{4},y_{3}y_{4}y_{5}\in J$ but $y_{1}y_{4}y_{5}$ or $y_{2}y_{4}y_{5}$ are not elements of $J$ and this is a contradiction.

**4): **

Let $\gcd(J_{1})=1=\gcd(J_{2})$. Suppose that $J_{1}$ is a square-free Veronese ideal and $J_{2}$ is an almost square-free Veronese ideal. We assume that $J_{2}=(y_{2}y_{3}y_{4},y_{2}y_{3}y_{5},y_{2}y_{3}y_{6},y_{2}y_{4}y_{5},y_{2}y_{4}y_{6})$. Since $\mid G(J_{3})\mid=3$, we can assume that one of the element $y_{3}y_{5}y_{6}$ or $y_{3}y_{4}y_{6}$ are not in $J$. If $y_{3}y_{5}y_{6}\notin J$, then $y_{2}y_{3}y_{5},y_{1}y_{5}y_{6}\in J$ but $y_{2}y_{5}y_{6}$ or $y_{3}y_{5}y_{6}$ are not elements of $J$ and this is a contradiction. If $y_{3}y_{4}y_{6}\notin J$, then $(J:y_{6})=(y_{1}y_{2},y_{1}(y_{3},y_{4},y_{5}),y_{2}(y_{3},y_{4}),y_{3}y_{5},y_{4}y_{5})$. Therefore by using Theorem 3.8 this is not SCM. Thus we do not have this case. Also, by the same argument of the Case (i), $J_{1}$ and $J_{2}$ are not almost square-free Veronese ideals in the same time. Therefore $J_{1}$, $J_{2}$ are square-free Veronese ideals and we have the case $(b)$.

**Case (iii): **

Let $\mid G(J_{3})\mid=2$. Then by Lemmas 4.6, 4.8, we have $\gcd(J_{1})=y_{3}$, $\gcd(J_{2})=1$ or $\gcd(J_{1})=1=\gcd(J_{2})$. If $\gcd(J_{1})=y_{3}$, $\gcd(J_{2})=1$, then we can assume that $G(J_{3})=\{y_{3}y_{4}y_{5},y_{3}y_{4}y_{6}\}$. Since $\gcd(J_{2})=1$, by Proposition 4.3 $J_{2}$ is square-free Veronese ideal or almost Veronese ideal. If $J_{2}$ is square-free Veronese ideal, then $y_{2}y_{5}y_{6},y_{3}y_{4}y_{5}\in J$ but $y_{3}y_{5}y_{6}$ or $y_{4}y_{5}y_{6}$ are not elements of $J$ and this is a contradiction. Let $J_{2}$ be an almost square-free Veronese ideal and we assume that $y_{5}y_{6}$ is the only element which is not in $J_{2}$. In this case by change of variables we have $J_{3}=0$ and $J_{1}=J_{2}$ are almost square-free Veronese ideals and and this is the case $(c)$. If $y_{4}y_{5}$ is the only element which is not in $J_{2}$, then $y_{3}y_{4}y_{5},y_{2}y_{4}y_{6}$ are elements of $J$ but $y_{2}y_{4}y_{5}$ or $y_{4}y_{5}y_{6}$ are not elements of $J$ and this is a contradiction. Also, if $y_{4}y_{6}$ is the only element which is not in $J_{2}$, then again $J$ is not matroidal and this is a contradiction. Now we can assume that $J_{3}=0$. If $\gcd(J_{1})=y_{3}$, then by Lemmas 4.6, 4.8 we have $\gcd(J_{2})=1$ or $\gcd(J_{2})=y_{3}$. If $\gcd(J_{2})=1$, then $y_{1}y_{3}y_{5}$ and $y_{2}y_{i}y_{j}$ are elements of $J$ for some $i,j=4,5,6$, but $y_{1}y_{i}y_{j}$ or $y_{3}y_{i}y_{j}$ are not elements of $J$ and this is a contradiction. Therefore $\gcd(J_{2})=y_{3}$ and this is the case $(c)$. Also, if $\gcd(J_{1})=1$ then $\gcd(J_{2})=1$. If $J_{1}\neq J_{2}$ are almost square-free Veronese ideals, then again by using the above argument $J$ is not matroidal and this is a contradiction. Therefore $J_{1}=J_{2}$ are square-free Veronese ideals or almost square-free Veronese ideals.

**Case (iv): **

Let $\mid G(J_{3})\mid=1$. Then by Lemmas 4.6, 4.8, we have $\gcd(J_{1})=1=\gcd(J_{2})$. Therefore by Proposition 4.3 $J_{1}$ and $J_{2}$ are square-free Veronese ideals or almost Veronese ideals. By choosing one element from $J_{1}$ and the only element from $J_{3}$, we have $\mid G(J_{3})\mid\geq 2$. This is a contradiction.