$d$-sequence and Regular sequence of Quadrics

TL;DR

This paper explores the algebraic structure of ideals generated by 1x1 minors of matrix products, showing how $d$-sequences and regular sequences naturally appear and using this to determine the Rees algebra's defining equations.

Contribution

It demonstrates the emergence of $d$-sequences and regular sequences in generators of specific ideals and computes the defining equations of their Rees algebra.

Findings

Identification of $d$-sequences and regular sequences in ideal generators

Calculation of the Rees algebra's defining equations

Application to ideals generated by 1x1 minors of matrix products

Abstract

Let $K$ be a field and $X$, $Y$ denote matrices such that, the entries of $X$ are either indeterminates over $K$ or $0$ and the entries of $Y$ are indeterminates over $K$ which are different from those appearing in $X$. We consider ideals of the form $I_{1}(XY)$, which is the ideal generated by the homogeneous polynomials of degree $2$ given by the $1\times 1$ minors of the matrix $XY$. We prove that $d$-sequences and regular sequences arise naturally as part of generators of $I_{1}(XY)$ for some special cases. We use this information to calculate the equations defining the Rees algebra of $I_{1}(XY)$.

Equations

$$(⟨a_0,\dots ,a_i⟩:a_{i+1}{a}_j)=(⟨a_0,\dots ,a_i⟩:a_j)\forall 0\le i\le n-1,\forall j\ge i+1.$$

$${\tilde{X}}_j=\left(\begin{array}{ccccc}{x}_{1j}& x_{11}& x_{12}& \cdots & x_{1n}\\ x_{2j}& x_{21}& x_{22}& \cdots & x_{2n}\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ x_{(n+1)j}& x_{(n+1)1}& x_{(n+1)2}& \cdots & x_{(n+1)(n+1)}\end{array}\right).$$

$$\{T_i{g}_j-T_j{g}_i,\sum _{i=1}^{n+1}-(1{)}^k{\Delta }_k{T}_k\mid 1\le i,j\le n+1\}.$$

$$(⟨g_0,\dots ,g_i⟩:g_{i+1}{g}_{n+1})=(⟨g_0,\dots ,g_i⟩:g_{n+1})$$

$${\delta }_{1n}=\left(\begin{array}{ccc}{X}_{11}& \cdots & X_{1n}\\ ⋮& ⋮& ⋮\\ X_{n1}& \cdots & X_{nn}\end{array}\right),d_{1n}=\left(\begin{array}{ccc}{g}_1& \cdots & g_n\end{array}\right),{\varphi }_{1n}=\left(\begin{array}{ccc}{y}_1& \cdots & y_n\end{array}\right).$$

$$\cdots ⟶(R(-4){)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}\oplus (R(-n-1){)}^{\left(\genfrac{}{}{0ex}{}{n}{1}\right)}\stackrel{{\alpha }_{2n}}{⟶}R(-2{)}^n\oplus R(-n)\stackrel{{\alpha }_{1n}}{⟶}R⟶R/T_n⟶0$$

$$\gamma_{1n}=\left[\begin{array}[]{c|c}-g_{n+1}I_{n}&B_{n}\\ \hline\cr d_{1n}&(-1)^{n+1}\Delta_{n+1}\end{array}\right],\quad\mbox{with}\quad B_{n}=\begin{pmatrix}-\Delta_{1}\\ \vdots\\ (-1)^{n}\Delta_{n}\end{pmatrix}.$$

$$\cdots ⟶(R(-4){)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)+n}\oplus R(-n-2)\stackrel{{\psi }_{2n}}{⟶}R(-2{)}^{n+1}\stackrel{{\psi }_{1n}}{⟶}R⟶R/I_1(XY)⟶0$$

$$X_{n}=\left[\begin{array}[]{ccccc}0&x_{12}&x_{13}&\ldots&x_{1n}\\ -x_{12}&0&x_{23}&\ldots&x_{2n}\\ -x_{13}&-x_{23}&&\\ \vdots&\vdots&&\\ -x_{1n}&-x_{2n}&\ldots&&0\end{array}\right]\quad\mbox{and}\quad Y_{n}=\left[\begin{array}[]{c}y_{1}\\ \vdots\\ y_{n}\end{array}\right].$$

$$C_n:=(I_n:J_n)=⟨g_{1(n-1)},g_{2(n-2)},\cdots ,g_{n-1(n-1)},y_n,{\Delta }_{(n)n}⟩.$$

$$B_n:=⟨g_{1(n-1)},g_{2(n-2)},\cdots ,g_{n-1(n-1)},y_n,{\Delta }_{(n)n}⟩$$

$$P_n:=(⟨g_{1(n-1)},g_{2(n-2)},\cdots ,g_{(n-1)(n-1)}⟩:{\Delta }_{(n)n})=⟨y_1,\cdots ,y_{n-1}⟩.$$

$$(⟨g_{0n},g_{1n},\cdots ,g_{in}⟩:g_{(j+1)n}{g}_{nn})=⟨(g_{0n},g_{1n},\cdots ,g_{in}⟩:g_{nn}),$$

$$(⟨g_{0n},g_{1n},\cdots ,g_{in}⟩:g_{nn})\subseteq (⟨g_{0n},g_{1n},\cdots ,g_{in}⟩:g_{(j+1)n}{g}_{nn}).$$

$$I_1(X_n{Y}_n)=⟨g_{1n},\dots ,g_{nn},{\Delta }_n⟩\cap ⟨y_1,\dots ,y_n⟩$$

$${\tilde{h}}_n\cdot p=\sum _{i=1}^{n-1}{h}_i{p}_i+\sum _{i=1}^r{m}_i(q_i-b_{k_i}).$$

$$g\cdot p^{\prime }=\sum _{i=1}^{n-1}{h}_i{p}_i^{\prime }+\sum _{i=1}^r{m}_{k_i}(q_i^{\prime }+b_{k_i})+{\tilde{h}}_n{p}_n^{\prime }.$$

$$\mathcal{F}=\{f_{11},\dots ,f_{n1}\}\cup \{f_{12},f_{32},\dots ,f_{k_{2}2}\}\cup \{f_{13},f_{43},\dots ,f_{k_{3}3}\}\cup \cdots \cup \{f_{1n}\}$$

$$\left(\begin{array}{cccccc}{f}_{11}& f_{12}& f_{13}& f_{14}& \cdots & f_{1n}\\ f_{21}& \times & \times & \times & \cdots & \times \\ f_{31}& f_{32}& \times & \times & \cdots & \times \\ f_{41}& \times & f_{43}& \times & \cdots & \times \\ f_{51}& f_{52}& \times & f_{54}& \cdots & \times \\ f_{61}& \times & \times & \times & \cdots & \times \\ f_{71}& f_{72}& f_{73}& \times & \cdots & \times \\ f_{81}& \times & \times & \times & \cdots & \times \\ ⋮& ⋮& ⋮& ⋮& \cdots & ⋮\end{array}\right)$$

$$x_{11}\cdots >x_{nn}$$

$$\tilde{\mathcal{F}}$$

Full text

The author is supported by the NPDF fellowship PDF/2019/001074, sponsored by the SERB, Government of India.

The second author is the corresponding author; supported by the MATRICS research grant MTR/2018/000420, sponsored by the SERB, Government of India.

1. Introduction

Our study originated from the 1974 paper of J. Herzog [1] on the following theme: Let $R$ be a Noetherian commutative ring with identity. Let $\underline{x}=\{x_{1},\ldots,x_{n}\}$ be a sequence in $R$. Let $\mathfrak{a}=(\alpha_{ij})$ be an $m\times n$ matrix with entries in $R$; with $m\leq n$. A complex $D_{*}(\underline{x},\mathfrak{a})$ was constructed in [1]. Acyclicity conditions on the complex $D_{*}(\underline{x},\mathfrak{a})$ were derived in order to decide perfectness and the Gorenstein property for the ideals $\langle a_{1},\ldots,a_{n},\Delta\rangle$ and $\langle a_{1},\ldots,a_{n},\Delta_{1},\ldots,\Delta_{n}\rangle$, where $a_{i}=\sum_{j=1}^{n}\alpha_{ij}x_{j}$, $i=1,\ldots,n$, $\Delta=\det(\alpha_{ij})$ (when $n\geq 2$ and $m=n$) and $\Delta_{j}$ is the determinant of the matrix obtained from $\mathfrak{a}$ by deleting the $j$-th column (when $n\geq 3$ and $m=n-1$). Our aim in this paper is to study a class of ideals of the form $I_{1}(XY)$ defined through determinantal conditions, which is similar to the aforesaid class studied by Herzog. We show that $d$-sequences and regular sequences occur very naturally in the setting of the ideals $I_{1}(XY)$, which is the principal object of study in this paper. We show how $d$-sequences occur naturally in Theorems 3.2 and 3.7. In Theorem 4.3 we show the occurrence of a regular sequence which follows an interesting pattern. However, this regular sequence is not a maximal one and the question finding one of maximal length remains open.

Ideals of the form $I_{1}(XY)$ have appeared in another significant work by Huneke and Ulrich [2]. We will use some of their observations for proving some results on $d$-sequences when $X$ is alternating, in Theorem 3.7. In this process we would also prove the conjectures proposed in the paper [6].

A sequence of (homogeneous) polynomials $p_{1},p_{2},\ldots,p_{r}$ in a polynomial ring $R$ is called a regular sequence if the ideal $\langle p_{1},p_{2},\ldots,p_{r}\rangle\neq R$ and if each $p_{i}$ is non-zero divisor in $R/\langle p_{1},p_{2},\ldots,p_{i-1}\rangle$, for every $1\leq i\leq r$. The notion of a $d$-sequence was defined by Huneke (see def 1), which generalizes the notion of a regular sequence.

Let Given an ideal $I=\langle a_{1},\ldots,a_{n}\rangle$ in a Noetherian ring $R$, the Rees algebra of $I$ is the $R$-algebra $R[It]=R[a_{1}t,\ldots,a_{n}t]$. One can define an $R$-algebra homomorphism $\pi:R[t_{1},\ldots,t_{n}]\rightarrow R[It]$ as $\pi(t_{i})=a_{i}t$. The map $\pi$ is a graded map of degree [math]. Therefore $\ker(\pi)$ is generated by homogeneous polynomials in $t_{1},\ldots,t_{n}$. Generators of $\ker(\pi)$ are called the equations defining the Rees algebra $R[It]$. Note that the linear polynomials $f_{ij}=a_{i}t_{j}-a_{j}t_{i}$ belong to $\ker(\pi)$ and we say that the ideal $I$ is of linear type (a notion introduced by Valla) if $\ker(\varphi)$ is generated by the linear homogeneous polynomials $f_{ij}$. It was proved independently by C. Huneke and G. Valla, and later generalized by K.N. Raghavan, that, if an ideal is generated by a $d$-sequence then the ideal is of linear type.

2. The ideal $I_{1}(XY)$

Let $R=K[x_{ij},y_{ij}\mid 1\leq i,j\leq n]$, $n\geq 2$, and $x_{ij}$ and $y_{ij}$ are distinct indeterminates over the field $K$. Let $X=(x_{ij})_{n\times n}$ and $Y=(y_{ij})_{n\times n}$ be generic matrices. Let us write the product of the matrices $X$ and $Y$ as $XY=(f_{ij})_{n\times n}$, so that $f_{ij}=\sum_{k=1}^{n}x_{ik}y_{kj}$. Let $I_{1}(XY)$ denote the ideal generated by the polynomials $f_{ij}$, which are the $1\times 1$ minors of the matrix $XY$. Certain properties like primality, primary decomposition and minimal free resolutions have been studied in [5], [4], [3].

It is easy to see that all the $f_{ij}$’s defined above do not form a regular sequence. For example, if $n=2$ then $x_{12}y_{21}f_{11}+x_{11}y_{12}f_{12}-x_{22}y_{21}f_{21}=x_{21}y_{12}f_{22}$ shows that $f_{11},f_{12},f_{21},f_{22}$ is not a regular sequence. However, certain interesting and useful results proved in this direction are the following:

Theorem 2.1**.**

Let $S=K[x_{ij},\,y_{j}\mid 1\leq i,j\leq n]$ denote the polynomial $K$-algebra and $T_{1},\ldots,T_{n}$ are indeterminates over $S$. Suppose that $X=(x_{ij})_{n\times n}$ is either generic or generic symmetric and $Y=(y_{j})_{n\times 1}$ is generic. Let $\mathcal{I}=I_{1}(XY)=\langle g_{1},\ldots,g_{n}\rangle$, where $g_{i}=\sum_{j=1}^{n}x_{ij}y_{j}$. The set $\{g_{1},\cdots,g_{n}\}$ forms a regular sequence. Hence, the defining equations of the Rees algebra of $\mathcal{I}$ are only the Koszul relations $T_{i}g_{j}-T_{j}g_{i}$, for $1\leq i<j\leq n$.

Proof.

See Theorem 6.1 in [5].∎

Theorem 2.2**.**

Let $S=K[x_{ij},\,y_{j}\mid 1\leq i,j\leq n]$ denote the polynomial $K$-algebra and $X=(x_{ij})_{n\times n}$ and $Y=(y_{j})_{n\times 1}$ be generic matrices. Suppose that $\mathcal{I}=I_{1}(XY)=\langle g_{1},\ldots,g_{n}\rangle$, where $g_{i}=\sum_{j=1}^{n}x_{ij}y_{j}$. Let us define the ideals $I_{i}=\langle g_{1},\ldots,g_{i}\rangle$, for $1\leq i\leq n-1$, and $I_{n}=\langle g_{1},\ldots,g_{n},\Delta\rangle$, where $\Delta=\det(X)$. The ideals $I_{1},\ldots,I_{n-1}$ and $I_{n}$ are all prime ideals.

Proof.

See Corollary 3.2, Theorem 3.3 and Theorem 4.4 in[4]. ∎

3. $d$-sequence

Definition 1**.**

[Definition 5.5.2; [7]] Let $A$ be a commutative ring. Set $a_{0}=0$. A sequence of elements $a_{1},\ldots,a_{n}$ is said to be a $d$-sequence if

[TABLE]

Let $S=K[x_{ij},\,y_{j}\mid 1\leq i\leq n+1,1\leq j\leq n]$ denote the polynomial $K$-algebra and $\widehat{X}=(x_{ij})_{(n+1)\times n}$, $Y=(y_{j})_{n\times 1}$ be generic matrices. Suppose $\mathcal{J}=I_{1}(\widehat{X}Y)=\langle g_{1},\ldots,g_{n+1}\rangle$, where $g_{i}=\sum_{j=1}^{n}x_{ij}y_{j}$. We now prove Theorem 3.2 to show that $g_{1},\ldots,g_{n+1}$ is a $d$-sequence. This would make the ideal an ideal of linear type and the equations defining its Rees algebra are all linear. In other words, the Rees algebra and the Symmetric algebra of the ideal are isomorphic. We prove the following lemma first:

Lemma 3.1**.**

Let $S$, $\widehat{X}$, $Y$ and $I_{1}(\widehat{X}Y)$ be as above. Let $\Delta_{i}$ denote the determinant of the matrix obtained by deleting $i$’th row of $\widehat{X}$ for $1\leq i\leq n+1$. Then,

1. (i)

$\displaystyle\sum_{i=1}^{n+1}(-1)^{i}x_{ij}\Delta_{i}=0$,  for all $j\in\{1,\ldots,n\}$. 2. (ii)

$\displaystyle\sum_{i=1}^{n+1}(-1)^{i}g_{i}\Delta_{i}=0$. 3. (iii)

$(\langle g_{1},\ldots,g_{n}\rangle:g_{n+1})=\langle g_{1},\ldots,g_{n},\Delta_{n+1}\rangle$. 4. (iv)

$(\langle g_{1},\ldots,g_{n}\rangle:\Delta_{n+1})=\langle y_{1},\ldots,y_{n}\rangle$.

Proof.

(i)   For each $1\leq j\leq n$, we consider the $(n+1)\times(n+1)$ matrix,

[TABLE]

Then we have $\det(\widetilde{X}_{j})=0$ for all $1\leq j\leq n$. Expanding with respect to the first column, we get $\displaystyle\sum_{i=1}^{n+1}(-1)^{i}x_{ij}\Delta_{i}=0$,  for all $j\in\{1,\ldots,n\}$.

(ii)  Rearranging terms we get, $\displaystyle\sum_{i=1}^{n+1}(-1)^{i}g_{i}\Delta_{i}=\displaystyle\sum_{i=1}^{n}(\displaystyle\sum_{i=1}^{n+1}(-1)^{i}x_{ij}\Delta_{i})y_{j}$. Therefore $\displaystyle\sum_{i=1}^{n+1}(-1)^{i}g_{i}\Delta_{i}=0$.

(iii) follows from Lemma 6.7 in [5].

(iv) By lemma 4.3 in [4], we have $\langle y_{1},\ldots,y_{n}\rangle\subset(\langle g_{1},\ldots,g_{n}\rangle:\Delta_{n+1})$. Let $h\Delta_{n+1}\in\langle g_{1},\ldots,g_{n}\rangle\subset\langle y_{1},\ldots,y_{n}\rangle$. Since $\langle y_{1},\ldots,y_{n}\rangle$ is a prime ideal, we have $h\in\langle y_{1},\ldots,y_{n}\rangle$.∎

Theorem 3.2**.**

Let $S=K[x_{ij},\,y_{j}\mid 1\leq i\leq n+1,1\leq j\leq n]$ denote the polynomial $K$-algebra and $\widehat{X}=(x_{ij})_{(n+1)\times n}$, $Y=(y_{j})_{n\times 1}$ be generic matrices. Suppose $\mathcal{J}=I_{1}(\widehat{X}Y)=\langle g_{1},\ldots,g_{n+1}\rangle$, where $g_{i}=\sum_{j=1}^{n}x_{ij}y_{j}$. Let $\Delta_{i}$ denote the determinant of the matrix obtained by deleting $i$’th row of $\widehat{X}$ for $1\leq i\leq n+1$.

1. (i)

The sequence $g_{1},\ldots,g_{n+1}$ forms a $d$-sequence. 2. (ii)

The Rees algebra of $I_{1}(\widehat{X}Y)$ is isomorphic to $S[T_{1},\ldots,T_{n+1}]/\mathfrak{I}$, where $\mathfrak{I}\subset S[T_{1},\ldots,T_{n}]$ is the ideal generated by the set

[TABLE]

Proof.

(i)   It is known by Theorem 2.1 that $g_{1},\ldots,g_{n}$ is a regular sequence. Every regular sequence is also a $d$-sequence, therefore it is enough to prove that

[TABLE]

for all $0\leq i\leq n$, where $g_{0}=0$. By Lemma 6.7 in [5], we have, $(\langle g_{0},\ldots,g_{n}\rangle:g_{n+1})=\langle g_{1},\ldots,g_{n},\Delta_{n+1}\rangle$. Again by Theorem 2.2, the ideals $\langle g_{0},\ldots,g_{i}\rangle$ for $1\leq i\leq n-1$ and $\langle g_{1},\ldots,g_{n},\Delta_{n+1}\rangle$ are prime ideals. Therefore, we have $(\langle g_{0},\ldots,g_{i}\rangle:g_{i+1}g_{n+1})=(\langle g_{0},\ldots,g_{i}\rangle:g_{n+1})$ for all $0\leq i\leq n$.

(ii)   A graded free minimal resolution of $I_{1}(\widehat{X}Y)$ can be found in Section 6 of [5], the construction of which uses the mapping cone technique. We now show, how this piece of information can be used to write the first syzygies matrix because no cancellation takes place at that level. We will be able to write the equations defining the Rees algebra if we can write the first syzygy matrix explicitly.

Let $P_{n}=\langle y_{1},\ldots,y_{n}\rangle$, $L_{n}=\langle g_{1},\ldots,g_{n}\rangle$, $T_{n}=\langle g_{1},\ldots,g_{n},\Delta_{n+1}\rangle$. Let $X_{ij}$ denote the $(ij)$-th cofactor of the matrix $X=(x_{ij})_{n\times n}$. We have the following diagram of exact chain complexes:

\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(R(-n-2))^{\binom{n}{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{2n}}$$\scriptstyle{\delta_{2n}}$$\textstyle{(R(-n-1))^{\binom{n}{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{1n}}$$\scriptstyle{\delta_{1n}}$$\textstyle{R(-n)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\phi_{0n}}$$\scriptstyle{\delta_{0n}=\Delta_{n}}$$\textstyle{R/P_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(R(-4))^{\binom{n}{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}n}$$\textstyle{(R(-2))^{\binom{n}{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}n}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{0n}}$$\textstyle{R/L_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}

where

[TABLE]

By Lemma 6.5 in [5], the first box in the above diagram is commutative. It is evident that, for each $0\leq i\leq n$, the degrees of the domain and the co-domain of $\delta_{in}$ do not match. Therefore, a minimal graded free resolution of $T_{n}$ is obtained by the mapping cone technique:

[TABLE]

where $\alpha_{1n}=\begin{pmatrix}g_{1}&\cdots&g_{n}&\Delta_{n}\end{pmatrix}$ and \,\alpha_{2n}=\left[\begin{array}[]{c|c}-\delta_{1n}&d_{2n}\\ \hline\cr\phi_{1n}&0\end{array}\right]. We now have the following diagram of exact chain complexes:

\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(R(-6))^{\binom{n}{2}}\oplus(R(-n-3))^{\binom{n}{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{2n}}$$\scriptstyle{\gamma_{2n}}$$\textstyle{R(-4)^{n}\oplus R(-n-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{1n}}$$\scriptstyle{\gamma_{1n}}$$\textstyle{R(-2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{0n}}$$\scriptstyle{\gamma_{0n}=g_{n+1}}$$\textstyle{R/T_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{\cdots\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{(R(-4))^{\binom{n}{2}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{2}n}$$\textstyle{(R(-2))^{\binom{n}{1}}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{1}n}$$\textstyle{R\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{d_{0n}}$$\textstyle{R/L_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}

where

[TABLE]

By Lemma 3.1, the first box is commutative in the above diagram. Therefore, a graded free resolution of $I_{1}(XY)$ is given by:

[TABLE]

where $\psi_{1n}=\begin{pmatrix}g_{1}&\cdots&g_{n+1}\end{pmatrix}$ and \psi_{2n}=\left[\begin{array}[]{c|c}-\gamma_{1n}&d_{2n}\\ \hline\cr\alpha_{1n}&0\end{array}\right].

We know that the sequence $\{g_{1},\ldots,g_{n+1}\}$ is a $d$-sequence. Therefore, the ideal $I_{1}(\widehat{X}Y)$ is of linear type and the syzygy matrix \psi_{2n}=\left[\begin{array}[]{c|c}-\gamma_{1n}&d_{2n}\\ \hline\cr\alpha_{1n}&0\end{array}\right] describes the ideal $\mathfrak{I}$ explicitly.∎

3.1. $I_{1}(XY)$ with alternating $X$.

Given an $n\times n$ generic alternating matrix $X$ and a generic column matrix $Y$, a scheme was proposed for computing a minimal free resolution of the ideal $I_{1}(XY)$ in the paper [6]. However, the scheme depended on two conjectures. We now prove those conjectures in Lemma 3.5 and Lemma 3.6 below. This not only validates our earlier work in [6] but also helps us prove Theorem 3.7 on the $d$-sequence.

We use the same notations as in [6]. Let $X_{n}$ denote the $n\times n$ alternating matrix and $Y_{n}$ denote the $n\times 1$ generic matrix given by

[TABLE]

With the assumption that $x_{ij}=-x_{ji}$, if $i>j$ and $x_{ii}=0$, let $g_{ki}=\Sigma_{j=1}^{i}x_{kj}y_{j}$. Therefore $I_{1}(X_{n}Y_{n})=\langle g_{1n},g_{2n},\cdots,g_{nn}\rangle$. Let $\Delta_{(i)n}$ denote the Pffafian of the alternating matrix $X_{n}$ with the $i$-th row and the $i$-th column deleted.

Lemma 3.3**.**

Let $x_{ij}=-x_{ji}$, if $i>j$ and $x_{ii}=0$. Then

• (i)

$y_{n}g_{nn}=-\left(y_{1}g_{1n}+y_{2}g_{2n}+\cdots+y_{n-1}g_{(n-1)n}\right)$.

• (ii)

$g_{k(n-1)}g_{nn}=x_{kn}y_{1}g_{1n}+x_{kn}y_{2}g_{2n}+\cdots+g_{nn}+x_{kn}y_{k}g_{kn}+\cdots+x_{kn}y_{n}g_{n-1n}$.

• (iii)

$\Delta_{(i)n}y_{i}=(-\Delta_{(1)n})g_{1n}+(\Delta_{(2)n})g_{2n}+\cdots+\widehat{(-1)^{i-1}\Delta_{(i)n})g_{in}}+\cdots((-1)^{n-1}\Delta_{(n-1)n})g_{(n-1)n}$,   for   every   $1\leq i\leq n$.

Proof.

These are easy to verify. ∎

Lemma 3.4**.**

$\{g_{1n},\cdots,g_{(n-1)n}\}$* forms a regular sequence for $n\geq 2$.*

Proof.

See part (ii) of Theorem 2.3 in [4]. ∎

Lemma 3.5**.**

Let $I_{n}=\langle g_{1n},g_{2n},\cdots,g_{(n-1)n}\rangle$ and $J_{n}=\langle g_{nn}\rangle$, so that $I_{n}+J_{n}=I_{1}(X_{n}Y_{n})$. Let

[TABLE]

If $n$ is even, then, $\Delta_{(n)n}=0$ and $C_{n}=\langle g_{1(n-1)},g_{2(n-2)},\cdots,g_{(n-1)(n-1)},y_{n}\rangle$, for every $n\geq 4$. Hence $C_{n}$ is a prime ideal for every $n\geq 4$.

Proof.

At first we note that the ideal

[TABLE]

is a prime ideal by Proposition 5.8 and Lemma 5.12 in [2]. Let $hg_{nn}\in I_{n}\subset B_{n}$. We have $h\in B_{n}$, since $g_{nn}\notin B_{n}$. Thus $C_{n}\subset B_{n}$. The other inclusion easily follows from Lemma 3.3. ∎

Lemma 3.6**.**

If $n$ is odd then $\Delta_{(n)n}\neq 0$, and for every $n\geq 4$,

[TABLE]

Proof.

Clearly $\langle y_{1},\cdots,y_{n-1}\rangle\subset P_{n}$ by the Lemma 3.3. Let $h\Delta_{(n)n}\in I_{n-1}\subset\langle y_{1},\cdots,y_{n-1}\rangle$. Since $\Delta_{(n)n}\notin\langle y_{1},\cdots,y_{n-1}\rangle$, we have $h\in\langle y_{1},\cdots,y_{n-1}\rangle$.∎

Theorem 3.7**.**

$\{g_{1n},\cdots,g_{(n-1)n},g_{nn}\}$* forms a $d$-sequence for $n\geq 3$.*

Proof.

We have $\{g_{1n},\cdots,g_{(n-1)n}\}$ forms a regular sequence for $n\geq 2$. Therefore, it is enough to show that, for $1\leq i\leq n-1$ and $j\geq i+1$,

[TABLE]

where $g_{0n}=0$ for all $n$. We have, for every $j\geq i+1$,

[TABLE]

Let $h\in(\langle g_{0n},g_{1n},\cdots,g_{in}\rangle:g_{(j+1)n}g_{nn})$, then $hg_{(j+1)n}g_{nn}\in\langle g_{0n},g_{1n},\cdots,g_{in}\rangle$. For $1\leq i\leq n-2$, the ideal $\langle(g_{0n},g_{1n},\cdots,g_{in}\rangle$ is a prime ideal (see Theorem 3.3 in [4]), therefore $hg_{nn}\in\langle g_{0n},g_{1n},\cdots,g_{in}\rangle$. Moreover, for $i=n-1$, the ideal $(\langle g_{0n},g_{1n},\cdots,g_{(n-1)n}\rangle:g_{nn})$ is a prime ideal by Lemma 3.5, therefore $h\in(\langle g_{0n},g_{1n},\cdots,g_{(n-1)n}\rangle:g_{nn})$. ∎

As an upshot of the aforesaid results, we get a primary decomposition of the ideal $I_{1}(X_{n}Y_{n})$ and somemore information regarding its normality.

Theorem 3.8**.**

The primary decomposition of

[TABLE]

where $\Delta_{n}$ is the pfaffian of the matrix $X_{n}$.

Proof.

Let $\mathfrak{Q}_{1n}=\langle g_{1n},\ldots,g_{nn},\Delta_{n}\rangle$ and $\mathfrak{Q}_{2n}=\langle y_{1},\ldots,y_{n}\rangle$. We note that $\mathfrak{Q}_{2n}$ is a prime ideal and by proposition 5.8 and lemma 5.12 in [2], $\mathfrak{Q}_{1n}$ is a prime ideal. Obviously $I_{1}(X_{n}Y_{n})\subset\mathfrak{Q}_{1n}\cap\mathfrak{Q}_{2n}$. Let $h\in\mathfrak{Q}_{1n}\cap\mathfrak{Q}_{2n}$, where $h=\displaystyle\sum_{k=1}^{n}a_{k}g_{kn}+b\Delta_{n}$. Then $b\Delta_{n}\in\mathfrak{Q}_{2n}$. Hence $b\in\mathfrak{Q}_{2n}$. By lemma 3.3 $b\Delta_{n}\in I_{1}(X_{n}Y_{n})$, therefore $h\in I_{1}(X_{n}Y_{n})$. ∎

Corollary 3.9**.**

The ideal $I_{1}(X_{n}Y_{n})$ is a radical ideal.

Proof.

Follows from Theorem 3.8.∎

Definition 2**.**

Let $R$ be a Noetherian ring and $I$ an ideal. Then $I$ is normally torsionfree if $\mbox{Ass}R/I=\mbox{Ass}R/I^{n}$ for $n\geq 1$.

Theorem 3.10**.**

Let $R$ be a regular local ring and $I$ a reduced ideal. If $I$ is normally torsionfree, then $I$ is normal.

Proof.

See the Proposition 1.54 in [8].∎

Lemma 3.11**.**

The ideal $I_{1}(X_{n}Y_{n})$ is normal.

Proof.

Follows from Theorems 3.8 and 3.10.∎

4. Construction of a regular sequence using $I_{1}(XY)$

Lemma 4.1**.**

Let $h_{1},h_{2}\cdots,h_{n}\in R$ be such that with respect to a suitable monomial order on $R$ the leading terms of them are mutually coprime. Then, $h_{1},h_{2}\cdots,h_{n}$ is a regular sequence in $R$.

Proof.

. See lemma 2.2 in [4].∎

Lemma 4.2**.**

Let $h_{1},\ldots,h_{n-1}$ be distinct polynomials and for $1\leq r\leq n-1$ let $m_{1},\ldots m_{r+1}$ be distinct monomials in $R$. Suppose that the following properties are satisfied with respect to some monomial order on $R$:

1. (i)

${\rm Lt}(h_{i})=a_{i}b_{i}$* for every $1\leq i\leq n-1$;* 2. (ii)

$\gcd({\rm Lt}(h_{i}),{\rm Lt}(h_{j}))=1$* for every $1\leq i\neq j\leq n-1$;* 3. (iii)

$\gcd(m_{i},m_{j})=1$*   for every $1\leq i\neq j\leq r+1$;* 4. (iv)

$\gcd({\rm Lt}(h_{i}),m_{j})=1$*   for every $1\leq i\leq n-1$ and $1\leq j\leq r+1$.*

Let $h_{n}=b_{k_{1}}m_{1}+\ldots b_{k_{r}}m_{r}+m_{r+1}+(\textrm{lower order terms})$. Then

1. (1)

$h_{1},\ldots,h_{n-1},m_{1},\ldots m_{r},h_{n}$* is a regular sequence.* 2. (2)

Moreover, if $g$ is a polynomial such that $\gcd({\rm\mbox{Lt}}(g),h_{i})=1$ for $1\leq i\leq n-1$ and $\gcd({\rm\mbox{Lt}}(g),m_{i})=1$ for $1\leq i\leq r+1$, then $h_{1},\ldots,h_{n-1},m_{1},\ldots,m_{r},h_{n},g$ is a regular sequence.

Proof.

(1) The sequence $h_{1},\ldots,h_{n-1},m_{1},\ldots,m_{r}$ is a regular sequence by the Lemma 4.1. Let $\widetilde{h}_{n}=h_{n}-\sum_{i=1}^{r}b_{k_{i}}m_{i}$, then ${\rm\mbox{Lt}}(\tilde{h}_{n})=m_{r+1}$ is coprime with ${\rm\mbox{Lt}}(h_{1}),\ldots,{\rm\mbox{Lt}}(h_{n-1})$ and also coprime with $m_{1},\ldots,m_{r}$. Again by the Lemma 4.1, we can write $h_{1},\ldots,h_{n-1},m_{1},\ldots,m_{r},\widetilde{h}_{n}$ is a regular sequence.

Let $h_{n}\cdot p=\sum_{i=1}^{n-1}h_{i}p_{i}+\sum_{i=1}^{r}m_{i}q_{i}$, therefore

[TABLE]

This gives $\,p\in\langle h_{1},\ldots,h_{n-1},m_{1},\ldots,m_{r}\rangle$, since $h_{1},\ldots,h_{n-1},m_{1},\ldots,m_{r},\widetilde{h}_{n}$ is a regular sequence. Therefore, $h_{1},\ldots,h_{n-1},m_{1},\ldots m_{r},h_{n}$ is a regular sequence.

(2) Let $g\cdot p^{\prime}=\sum_{i=1}^{n}h_{i}p^{\prime}_{i}+\sum_{i=1}^{r}m_{i}q^{\prime}_{i}$, therefore

[TABLE]

Now $h_{1},\ldots,h_{n-1},m_{1},\ldots m_{r},\widetilde{h}_{n},g$   is a regular sequence by 4.1; hence $p^{\prime}\in\langle h_{1},\ldots,h_{n-1},m_{k_{1}},\ldots,m_{r},\tilde{h}_{n}\rangle$.∎

Theorem 4.3**.**

Suppose that $k_{t}=1+\left(\left[\frac{n}{t}\right]-1\right)t$, then $k_{n}=1$. The ordered set

[TABLE]

is a regular sequence in $R$. The polynomials occurring in the list follow the pattern indicated below:

[TABLE]

Proof.

Consider the lexicographic monomial order given by

[TABLE]

In order to show that the set $\mathcal{F}$ is a regular sequence we consider a larger ordered set of polynomials by adding some indeterminates in the list so that they follow the properties listed in Lemma 4.2. The new ordered set of polynomials we consider is

[TABLE]

If we can show that $\widetilde{\mathcal{F}}$ is a regular sequence in this order, then, under any permutation of this order the polynomials would still form a regular sequence because of homogeneity. Therefore, we can rearrange the entries of $\widetilde{\mathcal{F}}$ in such a way that the elements that appear first are the ones listed in $\mathcal{F}$ and that proves our claim. By Lemma 4.1, $f_{11},\ldots f_{n1},y_{12}$ is a regular sequence. Using Lemma 4.2 we can add $f_{12}=x_{11}y_{12}+x_{12}y_{22}+\sum_{k=3}^{n}x_{1k}y_{k2}$ in the list. Therefore, $f_{11},\ldots f_{n1},y_{12},f_{12}$ is a regular sequence. The indeterminate $y_{32}$ does not divide the any of the leading terms of $f_{11},\ldots f_{n1},y_{12},f_{12}-x_{11}y_{12}$. Therefore, $f_{11},\ldots f_{n1},y_{12},f_{12},y_{32}$ is a regular sequence, by Lemma 4.2. Therefore, $\widetilde{\mathcal{F}}$ is a regular sequence if we continue the same process. The above comment proves that $\mathcal{F}$ is a regular sequence.∎