An Algorithm for Determining Whether the Quotient Ring Modulo, an Ideal, Has a Cyclic Basis

Abstract

This paper focuses on the research of the existence of cyclic bases for the quotient ring modulo, a zero-dimensional ideal. Based on the necessary and sufficient condition for the existence of cyclic bases, it further deduces an equivalent condition for the existence of cyclic bases. In accordance with this condition, a new algorithm is proposed. The existence of a cyclic is determined by constructing a matrix and calculating its determinant in the algorithm. Compared with the original algorithm, it narrows the scope of the candidate element set and improves the computational efficiency of the algorithm.

1. Introduction

As a class of basis vector sets with cyclic structures, cyclic bases play a key role in the fields of mathematics and computer science. On the one hand, they can significantly simplify the representation form and analysis process of objects with cyclic characteristics [1]. On the other hand, they can effectively optimize the computational process and reduce the computational complexity in the research of coding theory. Cyclic bases also play an important role in the research of zero-dimensional ideals, especially in the field of solving rational univariate representations [2,3], and they also have significant applications in aspects such as the primary decomposition of ideals [3]. Therefore, the problem of solving the cyclic basis of the quotient ring of a zero-dimensional ideal is of great research significance.

Solving polynomial systems is a core problem in computational algebraic geometry, among which the classical symbolic methods include the Gröbner basis method [4], the resultant method [5], Wu’s method [6], and so on. In subsequent research, many scholars have made improvements to these algorithms. For example, in the Gröbner basis method, the FGLM [7] algorithm proposed by Faugère in 1990 converts the symbolic computation of Gröbner bases into linear algebra computations, and this research pioneered a new problem-solving approach. Additionally, the F4 [8] and F5 [9] algorithms proposed by Faugère have improved the efficiency of computing lexicographic Gröbner bases. In Wu’s method, Wang Dongming proposed the theory of regular systems [6], which is applied to triangular decomposition. In 2016, Wang Dongming established the connection between characteristic sets and lexicographic Gröbner bases [10]. Furthermore, based on this theory, many studies have focused on characteristic decomposition [11,12,13].

In recent years, a commonly used solution method has been the Rational Univariate Representation (RUR). Furthermore, in recent years, many scholars have devoted themselves to research in this field [14,15]. The Rational Univariate Representation (RUR) was developed from the Shape Lemma. In 1999, Rouillier [16] proposed a method for computing the Rational Univariate Representation (RUR) of a general zero-dimensional ideal I. Some of the methods proposed by Rouillier have laid an important foundation for subsequent research on related theories. In 2003, Noro and Yokoyama proposed an algorithm for computing RUR using the modular method [17]. In 2010, Tan Chang, Zhang Shugong [18] and others extended Rouillier’s algorithm to positive-dimensional ideals, making it possible to represent all zeros of the ideal I. In 2010, Zeng Guangxing and Xiao Shuijing [19] proposed computing RUR using Wu’s method, which results in a more concise univariate representation but loses the multiplicity information of solutions. When the quotient ring of the corresponding ideal has a cyclic basis, both the corresponding ideal and the quotient ring possess favorable properties. In 2017, Shang Baoxin proposed that the existence of a cyclic basis in the quotient ring of a zero-dimensional ideal is equivalent to the ideal having a Shape basis. Furthermore, by virtue of the idea of basis transformation, he provided the RUR for zero-dimensional ideals with a Shape basis [3]. In 2021, Pan Jian combined the Gröbner basis with the zero-width theory, and proposed the necessary and sufficient conditions for the existence of a cyclic basis in the quotient ring of a zero-dimensional ideal. Based on these conditions, he further presented an existence judgment algorithm for the cyclic basis of the quotient ring by traversing the candidate elements in the set and computing the minimal polynomial of its multiplication matrix using fast linear algebra methods [2].

The cyclic basis judgment algorithm in Reference [2] requires traversing the candidate elements in the set and calculating their minimal polynomials one by one to verify the existence of cyclic bases in the quotient ring. Moreover, the calculation steps for the minimal polynomial of each candidate element are independent of each other and need to be performed repeatedly. By investigating the relationship between the coefficients of candidate elements and the width of the ideal, this paper derives the equivalent condition for the existence of cyclic bases in the quotient ring and designs an algorithm that constructs a matrix using the coefficients of candidate elements to determine whether cyclic bases exist. Compared with the original method, the proposed method has a smaller scope of the candidate element set and higher computational efficiency. This paper reveals the relationship between the coefficients of candidate elements and the width of the ideal, providing new ideas and methods for the research of subsequent related issues.

2. Preliminaries

Most of the concepts used in this paper will be introduced firstly.

Let K be a field of characteristic zero and $K\left[X\right]$ be the polynomial ring in n variables, where $K\left[X\right]=K[x_1,x_2,\dots ,x_n]$. Let $\overline{K}$ be the algebraic closure of K. Let ${\mathbb{V}}_{\overline{K}}\left(I\right)=\{z\in {\overline{K}}^n\mid h\left(z\right)=0\mathrm{for}\mathrm{all}h\in I\}$ be the affine variety of the ideal I. Let T denotes the set of all monomials over $K\left[X\right]$, defined as $T=\{X^{\mathbf{\alpha }}=x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\dots x_n^{{\alpha }_n}|\mathbf{\alpha }=({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{N}}^n\}$. Let $\mathrm{lt}\left(h\right)$ denote the leading term of h.

Definition 1 

([5]). If a finite subset $G=\{g_1,g_2,\dots ,g_n\}$ of an ideal I satisfies

$$\langle \mathrm{lt}\left(G\right)\rangle \triangleq \langle \mathrm{lt}\left(g_1\right),\mathrm{lt}\left(g_2\right),\dots ,\mathrm{lt}\left(g_n\right)\rangle ,$$

then G is called a Gröbner basis of the ideal I with respect to the monomial order ≺.

Theorem 1 

([5]). Let $I\subset K[x_1,\dots ,x_n]$ be a zero-dimensional ideal, then the following are equivalent:

• ${\mathbb{V}}_K\left(I\right)$ is a finite set;
• $K[x_1,\dots ,x_n]/I$ is a finite-dimensional space over K;
• The K-vector space Span$\{u=x_1^{{\alpha }_1}{x}_2^{{\alpha }_2}\cdots x_n^{{\alpha }_n}\in T\left[X\right]|u\notin \langle lt\left(I\right)\rangle \}$ is finite-dimensional.

Definition 2 

([16]). A polynomial $t\in K\left[X\right]$ is called a separating element of ${\mathbb{V}}_{\overline{K}}\left(I\right)$ if for all $z_1,z_2\in {\mathbb{V}}_{\overline{K}}\left(I\right)$ with $z_1\ne z_2$, it holds that $t\left(z_1\right)\ne t\left(z_2\right)$.

Definition 3 

([20]). Let $I\subseteq K\left[X\right]$ be a zero-dimensional ideal, and let $N_1={({\omega }_1,{\omega }_2,\dots ,{\omega }_r)}^T$ be a basis vector of the monomial quotient ring with respect to the ordering ≺, where $dim\left(K\right[X]/I)=r$. For a given $h\in K\left[X\right]$, suppose

$$h{N}_1\equiv {(h{\omega }_1,h{\omega }_2,\dots ,h{\omega }_r)}^T\equiv M_h{N}_{1}modI,$$

Then $M_h\in K^{r\times r}$ is called the multiplication matrix of h in the quotient ring $K\left[X\right]/I$ with respect to $N_1$.

Proposition 1 

([16]). Let $M_h$ be the multiplication matrix of h in the quotient ring $K\left[X\right]/I$ with respect to $N_1$. For $\alpha \in {\mathbb{V}}_{\overline{K}}\left(I\right)$, let $\mu \left(\alpha \right)$ denote the multiplicity of α. Then the following hold:

• $Det\left(M_h\right)={\displaystyle \prod _{\alpha \in {\mathbb{V}}_{\overline{K}}\left(I\right)}}h{\left(\alpha \right)}^{\mu \left(\alpha \right)}$.
• $Tr\left(M_h\right)={\displaystyle \sum _{\alpha \in {\mathbb{V}}_{\overline{K}}\left(I\right)}}h\left(\alpha \right)\mu \left(\alpha \right)$.
• The characteristic polynomial of the matrix $M_h$ is ${\displaystyle \prod _{\alpha \in {\mathbb{V}}_{\overline{K}}\left(I\right)}}{(T-h\left(\alpha \right))}^{\mu \left(\alpha \right)}$.

Let t be a separating element. By constructing the multiplication matrix $M_t$ of t, it follows that each $t\left(z_i\right)$, where $z_i\in {\mathbb{V}}_{\overline{K}}\left(I\right)$, is an eigenvalue of the multiplication matrix.

This paper mainly studies problems in zero-dimensional ideals. Let $I=\langle f_1,f_2,\dots ,f_s\rangle$ be a zero-dimensional ideal in $K\left[X\right]$. By Theorem 1, it follows that ${\mathbb{V}}_{\overline{K}}\left(I\right)=\{z_1,z_2,\dots ,z_m\}$ is a finite set of points

3. Width of an Ideal

Let K be a field, $T\left(X\right)=T(x_1,x_2,\dots ,x_n)$ be the set of all monomials in the polynomial ring $K\left[X\right]$, where each monomial is of the form $u=x_1^{{\gamma }_1}{x}_2^{{\gamma }_2}\cdots x_n^{{\gamma }_n}$ with exponents $({\gamma }_1,{\gamma }_2,\dots ,{\gamma }_n)\in {\mathbb{N}}^n$($\mathbb{N}$ denotes the set of non-negative integers). ${\mathbb{V}}_{\overline{K}}\left(I\right)=\{{\xi }_1,{\xi }_2,\cdots ,{\xi }_m\}$. Define the operator:

$$\mathsf{\partial }u={\displaystyle \frac{1}{{\gamma }_1!{\gamma }_2!\cdots {\gamma }_n!}}·{\displaystyle \frac{{\mathsf{\partial }}^{{\gamma }_1+{\gamma }_2+\cdots +{\gamma }_n}}{\mathsf{\partial }{x}_1^{{\gamma }_1}\mathsf{\partial }{x}_2^{{\gamma }_2}\cdots \mathsf{\partial }{x}_n^{{\gamma }_n}}}.$$

This operator $\mathsf{\partial }u$ is called a differential operator, and its degree is defined as ${\sum }_{i=1}^n{\gamma }_i$. In particular, $\mathsf{\partial }1$ denotes the identity operator, where the monomial $1=x_1^0{x}_2^0\cdots x_n^0$.

Let $\mathcal{U}=\{\mathsf{\partial }u\mid u\in T(X\left)\right\}$. Then the shift operator ${\sigma }_{x_j}:\mathcal{U}\to \mathcal{U}\cup \left\{0\right\}$ is defined as follows:

$${\sigma }_{x_j}(\mathsf{\partial }u)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{{\gamma }_1!{\gamma }_2!\cdots ({\gamma }_j-1)!\cdots {\gamma }_n!}·\frac{{\mathsf{\partial }}^{{\gamma }_1+{\gamma }_2+\cdots +({\gamma }_j-1)+\cdots +{\gamma }_n}}{\mathsf{\partial }{x}_1^{{\gamma }_1}\mathsf{\partial }{x}_2^{{\gamma }_2}\cdots \mathsf{\partial }{x}_j^{{\gamma }_j-1}\cdots \mathsf{\partial }{x}_n^{{\gamma }_n}},}\hfill & \mathrm{if}{\gamma }_j\ge 1;\hfill \\ 0,\hfill & \mathrm{if}{\gamma }_j=0.\hfill \end{array}\right.$$

Let $({\eta }_1,{\eta }_2,\dots ,{\eta }_n)\in {\mathbb{N}}^n$ and $v=x_1^{{\eta }_1}{x}_2^{{\eta }_2}\cdots x_n^{{\eta }_n}\in T\left(X\right)$. By the relation ${\sigma }_{v{x}_i}={\sigma }_v{\sigma }_{x_i}$, the operator ${\sigma }_v:\mathcal{U}\to \mathcal{U}\cup \left\{0\right\}$ is defined. Denote

$${\overrightarrow{\mathrm{Span}}}_{\underline{K}}\left(\mathcal{U}\right):=\left\{\sum c_u\mathsf{\partial }u|c_u\in \underline{K},u\in \mathcal{U}\right\}.$$

where $\underline{K}$ is the smallest algebraic extension field containing ${\xi }_i$ for $1\le i\le m$.

The degree of an element in ${\overrightarrow{\mathrm{Span}}}_{\underline{K}}\left(\mathcal{U}\right)$ is the degree of the highest-degree differential operator $\mathsf{\partial }u$ within that element.

Definition 4 

([21]). Let $\Delta \subset {\overrightarrow{\mathrm{Span}}}_{\underline{K}}\left(\mathcal{U}\right)$ be a linear space. If Δ is finite-dimensional and is closed under the action of all shift operators ${\sigma }_v$, then Δ is called a closed subspace.

Definition 5 

([21]). Let $I\subset K\left[X\right]$ be a zero-dimensional ideal, and let z be a zero of I. Consider a family of differential operators $\mathcal{L}=\{L_0=\mathsf{\partial }1,L_1,\dots ,L_D\}$. Suppose these operators are linearly independent over the field K, and the subspace they span, ${\Delta }_z\left(I\right):={\overrightarrow{\mathit{Span}}}_{\underline{K}}(L_0,L_1,\dots ,L_D)$, is a closed subspace. Furthermore, if for all $L\in {\Delta }_z\left(I\right)$ and all $f\in I$, it holds that ${\left.L\left(f\right)\right|}_{X=z}=0$, then ${\Delta }_z\left(I\right)$ is called the closed subspace of differential operators at the zero z.

Let q be a non-negative integer, then ${\Delta }_z^{\left(q\right)}\left(I\right)$ is called the closed space of differential operators with degree not exceeding q in ${\Delta }_z\left(I\right)$.

Based on the above conditions, the definition of the width of a zero pint is as follows.

Definition 6 

([22]). Let $z\in {\mathbb{V}}_{\overline{K}}\left(I\right)$. The width κ of z is defined as:

$$\kappa =dim\left({\Delta }_z^{\left(1\right)}\left(I\right)\right)-dim\left({\Delta }_z^{\left(0\right)}\left(I\right)\right),0⩽\kappa ⩽n.$$

Furthermore, we can define the width of an ideal.

Definition 7 

([2]). The width of an ideal I is defined as the maximum width of the maximum width of the zero points of ideal I.

4. The Equivalent Condition for the Existence of a Cyclic Basis in the Quotient Ring Modulo a Zero-Dimensional Ideal

Focusing on a zero-dimensional ideal I, this section first puts forward the following conclusion: if the width of a zero-dimensional ideal I does not exceed 1, its quotient ring necessarily has a cyclic basis. Based on this, the main work of this section is to obtain the equivalence condition for the width of I being at most 1 by integrating the definition of width. This condition is also the equivalent condition for the existence of a cyclic basis in its quotient ring. This section will also prove the feasibility of this condition, which provides essential theoretical support for verifying the rationality of the algorithm in the following sections.

Lemma 1 

([23]). If $z\in {\mathbb{V}}_{\overline{K}}\left(I\right)$ satisfies $dim\left({\Delta }_z^{\left(1\right)}\left(I\right)\right)-dim\left({\Delta }_z^{\left(0\right)}\left(I\right)\right)=\kappa$, $1\le \kappa \le n$, then for any $t\in \mathbb{N}\setminus \left\{0\right\}$, the following holds:

$$dim\left({\Delta }_z^{\left(t\right)}\left(I\right)\right)-dim\left({\Delta }_z^{(t-1)}\left(I\right)\right)\le \left({\displaystyle \genfrac{}{}{0pt}{}{\kappa +t-1}{\kappa -1}}\right).$$

Based on the lemma, without loss of generality, we take the first variable $z_1$ for any $z\in {\mathbb{V}}_{\overline{K}}\left(I\right)$. If $dim\left({\Delta }_z^{\left(1\right)}\left(I\right)\right)-dim\left({\Delta }_z^{\left(0\right)}\left(I\right)\right)=1$, then for any $t\in \mathbb{N}\setminus \left\{0\right\}$, the following holds:

$$dim\left({\Delta }_z^{\left(t\right)}\left(I\right)\right)-dim\left({\Delta }_z^{(t-1)}\left(I\right)\right)\le 1.$$

Definition 8 

([2]). If the maximum width among the zeros of the ideal I is 1, then the ideal I is said to have width 1.

The following proposition is easily obtained.

Proposition 2. 

If the width of an ideal I is at most 1, then there exists $k\in K$ such that the multiplication matrix $M_t$ corresponding to $t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n$ is a non-defective matrix.

Corollary 1. 

If the multiplication matrix $M_t$ corresponding to any $t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n$, is a defective matrix, then the width of the ideal is at least 2.

Theorem 2 

([2]). Let $I\subseteq K\left[X\right]$ be a zero-dimensional ideal. The necessary and sufficient condition for the width of I to be at most 1 is that $K\left[X\right]/I$ has a cyclic basis.

From the above theorem, we can conclude that whether an quotient ring has a cyclic basis can be determined by verifying its width.

Let $N={({\omega }_1,{\omega }_2,\dots ,{\omega }_r)}^T$ be the monomial basis of the quotient ring $K\left[X\right]/I$. Then, for every $f\in K\left[X\right]$, its residue class $\left[f\right]$ modulo the ideal I can be expressed in terms of the monomial basis N. That is, there exist $a_i\in K$ for $i=1,\dots ,r$ such that $\left[f\right]=a_1{\omega }_1+\cdots +a_r{\omega }_r$. It is easily seen that the vector of $\left[f\right]$ with respect to the monomial basis is

$$\overrightarrow{\left[f\right]}=(a_1,a_2,\dots ,a_r).$$

For any $f^i\in K\left[X\right]$, we have $\left[f^i\right]\in K\left[X\right]/I$. Define $\overrightarrow{\left[f^i\right]}=(a_{i1},a_{i2},\dots ,a_{ir})$, where $\left[1\right]=(1,0,\dots ,0)$. For the convenience of subsequent notation in the paper, denote $\left[1\right]=(a_{01},a_{02},\dots ,a_{0r})$.

We now proceed to present the equivalent theorems for determining the width of an ideal.

Theorem 3. 

Let I be a zero-dimensional ideal. Let $t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n\in K\left[X\right]$ where $k\in \mathbb{Z}$. Denote the vector of $\left[t^i\right]$ with respect to the monomial basis as $\overrightarrow{\left[t^i\right]}=(a_{i1},a_{i2},\dots ,a_{ir})$. Then we have that the width of I is at least 2, if and only if, for all $k\in \mathbb{Z}$,$\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ are linearly dependent.

Proof. 

It is easily seen that, for all $k\in \mathbb{N}$, the linear dependence of $\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ is equivalent to the system of equations

$$\left\{\begin{array}{c}{a}_{01}{x}_1+a_{11}{x}_2+\cdots +a_{r-1,1}{x}_r=0,\hfill \\ a_{02}{x}_1+a_{12}{x}_2+\cdots +a_{r-1,2}{x}_r=0,\hfill \\ ⋮\hfill \\ a_{0r}{x}_1+a_{1r}{x}_2+\cdots +a_{r-1,r}{x}_r=0,\hfill \end{array}\right.$$

(1)

having a non-trivial solution for any $k\in \mathbb{Z}$.

“⇐” Let $N={({\omega }_1,{\omega }_2,\dots ,{\omega }_r)}^T$ be the monomial basis of the quotient ring $K\left[X\right]/I$. Since for $\forall k\in \mathbb{Z}$, the system of (1) has a non-trivial solution, denoted as $(b_0,\dots ,b_{r-1})$. Then for all $k\in \mathbb{Z}$, we have:

$$b_0·\overrightarrow{\left[1\right]}+b_1·\overrightarrow{\left[t\right]}+\cdots +b_{r-1}·\overrightarrow{\left[t^{r-1}\right]}=\overrightarrow{0}.$$

It follows that

$$\left(b_0·\overrightarrow{\left[1\right]}+b_1·\overrightarrow{\left[t\right]}+\cdots +b_{r-1}·\overrightarrow{\left[t^{r-1}\right]}\right)·{({\omega }_1,{\omega }_2,\dots ,{\omega }_r)}^T=0.$$

That is,

$$b_0+b_1·\left[t\right]+\cdots +b_{r-1}·\left[t^{r-1}\right]=0.$$

Since ${\sum }_{i=0}^{r-1}{b}_i\left[t^i\right]=0$, we have ${\sum }_{i=0}^{r-1}{b}_i{t}^i\in I$. Thus, for any monomial $x^{\alpha }\in K\left[X\right]$, it holds that

$$x^{\alpha }·\sum _{i=0}^{r-1}{b}_i{t}^i\in I.$$

In other words, $x^{\alpha }·{\sum }_{i=0}^{r-1}{b}_i{t}^i={\sum }_{i=0}^{r-1}{b}_i{x}^{\alpha }{t}^i\in I$, which implies ${\sum }_{i=0}^{r-1}{b}_i\left[x^{\alpha }{t}^i\right]=0$.

Taking $x^{\alpha }=1$, we obtain the following:

$$\sum _{i=0}^{r-1}{b}_i·1·t^i=\sum _{i=0}^{r-1}{b}_i\left[t^i\right]=0.$$

That is,

$$\sum _{i=0}^{r-1}{b}_i(1,0,0,\cdots ,0)M_{t^i}{({\omega }_1,\dots ,{\omega }_r)}^T=0,$$

which shows that the first row of ${\sum }_{i=0}^{r-1}{b}_i{M}_{t^i}$ is entirely zero.

Taking $x^{\alpha }={\omega }_2$, we obtain:

$$\sum _{i=0}^{r-1}{b}_i·{\omega }_2·t^i=\sum _{i=0}^{r-1}{b}_i\left[{\omega }_2{t}^i\right]=0.$$

That is,

$$\sum _{i=0}^{r-1}{b}_i(0,1,0,\cdots ,0)M_{t^i}{({\omega }_1,\dots ,{\omega }_r)}^T=0,$$

which shows that the second row of ${\sum }_{i=0}^{r-1}{b}_i{M}_{t^i}$ is entirely zero.

$⋮$

Taking $x^{\alpha }={\omega }_r$, we obtain:

$$\sum _{i=0}^{r-1}{b}_i·{\omega }_r·t^i=\sum _{i=0}^{r-1}{b}_i\left[{\omega }_r{t}^i\right]=0.$$

That is,

$$\sum _{i=0}^{r-1}{b}_i(0,0,\cdots ,1)M_{t^i}{({\omega }_1,\dots ,{\omega }_r)}^T=0,$$

which shows that the r-th row of ${\sum }_{i=0}^{r-1}{b}_i{M}_{t^i}$ is entirely zero.

In summary, ${\sum }_{i=0}^{r-1}{b}_i{M}_{t^i}$ is the zero matrix. Let $h\left(x\right)=b_0+b_1·x+\cdots +b_{r-1}·x^{r-1}$, then $h\left(M_t\right)=0$.

Let the minimal polynomial of the multiplication matrix $M_t$ be $h_{\alpha }\left(x\right)$. Then $h_{\alpha }\left(x\right)\mid h\left(x\right)$. It is easily seen that $deg\left(h_{\alpha }\left(x\right)\right)\le deg\left(h\left(x\right)\right)\le r-1$. Hence, the characteristic polynomial cannot be equal to the minimal polynomial, meaning the multiplication matrix $M_t$ is a defective matrix. By Corollary coro:!, the width of the ideal I is at least 2.

“⇒” Since the width of I is at least 2, $K\left[X\right]/I$ has no cyclic basis. This implies there does not exist any generator $t\in K\left[X\right]$ such that $\{1,t,t^2,\dots ,t^{r-1}\}$ forms a cyclic basis. In other words, there exist scalars $b_0,b_1,\dots ,b_{r-1}$, not all zero, that satisfy

$$b_0+b_1\left[t\right]+\cdots +b_{r-1}\left[t^{r-1}\right]=0.$$

That is,

$$(b_0+b_{1}t+\cdots +b_{r-1}{t}^{r-1})modI=0,$$

which implies:

$$b_0+b_1·\left[t\right]+\cdots +b_{r-1}·\left[t^{r-1}\right]=0.$$

Thus, we have $b_0·\overrightarrow{\left[1\right]}+b_1·\overrightarrow{\left[t\right]}+\cdots +b_{r-1}·\overrightarrow{\left[t^{r-1}\right]}=\overrightarrow{0}$.

This means the system of equations

$$\left\{\begin{array}{c}{a}_{01}{x}_1+a_{11}{x}_2+\cdots +a_{r-1,1}{x}_r=0,\hfill \\ a_{02}{x}_1+a_{12}{x}_2+\cdots +a_{r-1,2}{x}_r=0,\hfill \\ ⋮\hfill \\ a_{0r}{x}_1+a_{1r}{x}_2+\cdots +a_{r-1,r}{x}_r=0,\hfill \end{array}\right.$$

has a non-trivial solution for all $k\in \mathbb{Z}$.    □

By Theorem 3, we can derive the following corollary:

Corollary 2. 

The width of the ideal I is at most 1, if and only if, there exists $k\in \mathbb{Z}$ such that $\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ are linearly independent, where $t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n\in K\left[X\right]$.

Herein, we present equivalent conditions for the ideal I to have a width of at most 1. Since a width of at most 1 implies that the corresponding quotient ring has a cyclic basis, determining the width of the ideal is important for solving the problem. Subsequently, we provide a method for judging the width of the ideal based on the equivalent conditions.

Theorem 4. 

The width of the ideal I is at least 2 if, for all integers $k\in \mathbb{Z}$ where $0\le k\le r(r-1){\displaystyle \frac{(n-1)}{2}}$, the vectors $\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ are linearly dependent.

Proof. 

Denote the coefficient matrix of the system of (1) as:

$$A=\left(\begin{array}{cccc}{a}_{01}& a_{11}& \dots & a_{r-1,1}\\ a_{02}& a_{12}& \dots & a_{r-1,2}\\ ⋮& ⋮& \ddots & ⋮\\ a_{0r}& a_{1r}& \dots & a_{r-1,r}\end{array}\right).$$

By Theorem 3, the width of the ideal I being at least 2 is equivalent to the system (1) having a non-trivial solution for all $k\in \mathbb{N}$. From the existence theorem for solutions of homogeneous linear systems, $Ax=0$ has a non-trivial solution if and only if the rank of the coefficient matrix A is less than the number of unknowns, i.e., $\mathrm{rank}\left(A\right)<n$. This is equivalent to $\left|A\right|=0$. Since $\left|A\right|$ is an r-th order determinant, and by the earlier definition, $a_{01}=1$ and $a_{02}=\cdots =a_{0r}=0$, the first column of matrix A is ${(1,0,\dots ,0)}^T$. By properties of determinants, computing $\left|A\right|$ is equivalent to computing the determinant of the matrix:

$$B=\left(\begin{array}{ccc}{a}_{12}& \dots & a_{r-1,2}\\ ⋮& \ddots & ⋮\\ a_{1r}& \dots & a_{r-1,r}\end{array}\right).$$

For an $l\times l$ matrix $M=\left[m_{ij}\right]$, its determinant can be expressed as the algebraic sum over all permutations, by properties of determinants:

$$\mathrm{Det}\left(M\right)=\sum _{\sigma \in S_l}\mathrm{sgn}\left(\sigma \right)·m_{1\sigma \left(1\right)}{m}_{2\sigma \left(2\right)}\cdots m_{l\sigma \left(l\right)}.$$

where $S_l$ is the set of all permutations of $\{1,2,\dots ,l\}$, $\sigma$ denotes a permutation, $\sigma \left(i\right)$ indicates the column index of the element in row i, and $\mathrm{sgn}\left(\sigma \right)$ is the sign of the permutation ( $+1$ for even permutations, $-1$ for odd permutations). Since the determinant of a transpose equals the determinant of the original matrix, we may without loss of generality consider products of elements selected column-wise.

By the earlier definition, the degree of the function t with respect to k is $n-1$, so the degree of $\left[t\right]$ with respect to k is at most $n-1$. The degree of $t^2$ with respect to k is $2(n-1)$, so the degree of $\left[t^2\right]$ with respect to k is at most $2(n-1)$. Similarly, the degree of $t^i$ with respect to k is $i(n-1)$, so the degree of $\left[t^i\right]$ with respect to k is at most $i(n-1)$ for $0\le i\le r-1$. It follows that each term $a_{0\sigma \left(1\right)}{a}_{1\sigma \left(2\right)}\dots a_{r-1\sigma \left(r\right)}$ in the determinant sum has a maximum degree with respect to k of $r(r-1){\displaystyle \frac{(n-1)}{2}}$. Thus, the determinant itself has a maximum degree with respect to k of $r(r-1){\displaystyle \frac{(n-1)}{2}}$, so we can express $\mathrm{Det}\left(A\right)$ as a polynomial in k:

$$\mathrm{Det}\left(A\right)=b_{r(r-1){\displaystyle \frac{(n-1)}{2}}}{k}^{r(r-1){\displaystyle \frac{(n-1)}{2}}}+\cdots +b_{1}k+b_0.$$

where $b_0,b_1,\dots ,b_{r(r-1){\displaystyle \frac{(n-1)}{2}}}\in \mathbb{Z}$.

We conclude that $det\left(A\right)$ is a polynomial in k of degree $r(r-1){\displaystyle \frac{(n-1)}{2}}$, so $det\left(A\right)=0$ has at most $r(r-1){\displaystyle \frac{(n-1)}{2}}$ solutions. Therefore, if $g\left(k\right)=0$ holds for all k from 0 to $r(r-1){\displaystyle \frac{(n-1)}{2}}$,then the polynomial of the $\mathrm{Det}\left(A\right)$ with respect to the variable k is the zero polynomial, i.e., $\mathrm{Det}\left(A\right)=0$ identically. Hence, the width of the ideal I is at least 2.    □

By Corollary 2 and Theorem 4, we obtain the following theorem.

Theorem 5. 

The width of the ideal I is at most 1 if there exists $k\in \mathbb{Z}$ with $0\le k\le r(r-1){\displaystyle \frac{(n-1)}{2}}$ such that $\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ are linearly independent.

The theorem implies that if the determinant is non-zero when $k=j$ (where $0\le j\le r(r-1){\displaystyle \frac{(n-1)}{2}}$), then the width of the ideal I is at most 1. Conversely, if the determinant is zero for all k with $0\le k\le r(r-1){\displaystyle \frac{(n-1)}{2}}$, then the width of the ideal I is at least 2. Thus, it suffices to check all integers k from 0 to $r(r-1){\displaystyle \frac{(n-1)}{2}}$ to determine the width of the ideal I.

5. Algorithm for Determining the Width of an Ideal

By Theorem 5 from the previous section, if the width of the ideal is at most 1, then in the set

$$T=\left\{t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n\mid 0\le k\le r(r-1){\displaystyle \frac{(n-1)}{2}}\right\},$$

there exists at least one candidate element t such that the determinant of the matrix formed by its corresponding vectors is non-zero. Therefore, by determining whether there exists an element in the set T such that the determinant of the matrix formed by the vector corresponding to this element is non-zero, we can further judge whether the width of the ideal I is at most 1.

In the algorithm, since computing $\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ individually involves extensive calculations, the computation of these vectors can be simplified as follows.

Since $t·{\omega }_{1}modI=\left[t\right]$, we have $\overrightarrow{\left[t\right]}=(1,0,\dots ,0)·M_t$. Similarly, $\overrightarrow{\left[t^2\right]}=(1,0,\dots ,0)·M_{t^2}=(1,0,\dots ,0)·M_t·M_t$, which implies $\overrightarrow{\left[t^2\right]}=\overrightarrow{\left[t\right]}·M_t$. This leads to the recursive formula for the vectors:

$$\overrightarrow{\left[t^i\right]}=\overrightarrow{\left[t^{i-1}\right]}·M_t$$

It follows from Theorem 3 that if there exists $\left[t^i\right]$ which is a zero vector, then it can be directly concluded that the width of the ideal is at least 2. The following presents a method for determining the width of an ideal.

Algorithm 1: Algorithm for Determining the Width of an Ideal

Input: Ideal I, the monomial basis $N={({\omega }_1,{\omega }_2,\dots ,{\omega }_r)}^T$ of the quotient ring $K\left[X\right]/I$ with respect to the order ${\prec }_{\mathrm{grevlex}}$ and the multiplication matrix $M_{x_i}$ of $x_i$ with respect to N.

Output: The width of the ideal I.

Step 1: From the set $T=\{t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n\mid 0⩽k⩽(n-1){\displaystyle \frac{r(r-1)}{2}}\}$, compute $\overrightarrow{\left[1\right]},\overrightarrow{\left[t\right]},\dots ,\overrightarrow{\left[t^{r-1}\right]}$ one by one in the order of increasing k. If any vector is $\overrightarrow{0}$, select the next k and repeat Step 1; otherwise, proceed to Step 2.

Step 2: Form these vectors into an r-th order matrix $A={({\overrightarrow{\left[1\right]}}^T,{\overrightarrow{\left[t\right]}}^T,\cdots ,{\overrightarrow{\left[t^{r-1}\right]}}^T)}^T$. Remove the first row and the first column of A, and denote the matrix formed by the remaining elements as B.

Step 3: Compute Det$\left(B\right)$. If Det$\left(B\right)\ne 0$, terminate the computation and return that the width of the ideal I is at most 1.

Step 4: If for all $k\in \mathbb{N}$ where $0⩽k⩽(n-1){\displaystyle \frac{r(r-1)}{2}}$, $det\left(B\right)=0$ holds, then return that the width of the ideal I is at least 2.

Proposition 3. 

Let $I\subseteq K\left[X\right]$ be a zero-dimensional ideal, $dim\left(K\right[X]/I)=r$, and $T=\left\{t=x_1+k{x}_2+\cdots +k^{n-1}{x}_n\mid 0⩽k⩽(n-1){\displaystyle \frac{r(r-1)}{2}}\right\}$. Then the complexity of Algorithm 1 determines the Width of an Ideal is at most

$$O\left({\displaystyle \frac{5}{6}}n{r}^5\right)$$

basic arithmetic operations in K.

Proof. 

In complexity analysis, addition operations can be omitted since their order is lower than that of multiplication operations. Moreover, the most general case is considered in the analysis. From the formula $\overrightarrow{\left[t^i\right]}=\overrightarrow{\left[t^{i-1}\right]}·M_t,i=1,2,\cdots ,r-1$, the complexity of computing a single vector $\overrightarrow{\left[t^i\right]}$ requires $O\left(r^2\right)$ arithmetic operations in K. It follows that the complexity of Step 1 in Algorithm 1 needs $O\left(r^2(r-2)\right)=O\left(r^3\right)$ arithmetic operations in K.

For computing the determinant of an $(r-1)$-th order matrix, we first transform the matrix into an upper triangular matrix via row operations and then multiply the diagonal elements. The arithmetic operations of this process is at most $O\left({\displaystyle \frac{2r^3}{3}}\right)$.

That is, the arithmetic operations of determining whether the coefficients of a candidate element satisfy the existence condition of a cyclic basis is at most $O\left({\displaystyle \frac{5r^3}{3}}\right)$. Let $H_i$ denote the complexity of judging one candidate element in Algorithm 1. From the above analysis, it is known that computing $H_i$ needs at most $O\left({\displaystyle \frac{5r^3}{3}}\right)$ arithmetic operations in K. According to the most general case, the maximum number of steps requires iterating over $0\le k\le {\displaystyle \frac{(n-1)r(r-1)}{2}}$, and calculating the determinant value of the candidate element corresponding to each k to determine the width of the ideal. The total complexity of Algorithm 1 is $O\left({\sum }_{i=0}^{{\displaystyle \frac{(n-1)r(r-1)}{2}}}{H}_i\right)$.

When each $H_i$, $(i=1,2,\dots ,r)$ takes its maximum value, the expression simplifies to:

$$O\left({\displaystyle \frac{(n-1)r(r-1)}{2}}·{\displaystyle \frac{5r^3}{3}}\right)=O\left({\displaystyle \frac{5}{6}}n{r}^5\right).$$

That is, the complexity of Algorithm 1 is at most $O\left({\displaystyle \frac{5}{6}}n{r}^5\right)$. □

It follows from [2] that the complexity of the original algorithm is at most

$$O\left({\displaystyle \frac{3}{2}}n(r^5+r^{4}log\left(r\right)log(log\left(r\right)))\right).$$

From the complexities of the new algorithm and the original algorithm, it can be seen that the efficiency of the new algorithm is superior to that of the original algorithm.

Example 1. 

Consider the ideal $J=\langle x^2-yz,-xz+y^2,-xy+z^2,xyz-x,y^3-y\rangle$.

Using Maple 2024, we compute the Gröbner basis of the ideal J in the quotient ring $K[x,y,z]/J$ as $G=[x-y,-y^2+yz,-y^2+z^2,y^3-y]$, and its monomial basis as $(1,y,z,y^2)$. Thus, the set of candidate elements is $T=\{t=x+ky+k^{2}z\mid 0⩽k⩽12\}$.

We implement Algorithm 1 using Maple. When substituting $k=1$, we obtain $t=x+y+z$, and the corresponding multiplication matrix is

$$M_t=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 2& 1& 0\\ 0& 0& 0& 9\\ 0& 27& 0& 0\end{array}\right].$$

The corresponding matrix B is computed as

$$B=\left[\begin{array}{ccc}2& 1& 0\\ 0& 0& 9\\ 27& 0& 0\end{array}\right],$$

i.e., $\mathrm{Det}\left(B\right)\ne 0$.

The algorithm terminates at $k=1$, and we conclude that the width of the ideal J is at most 1, meaning there exists a cyclic basis in the quotient ring $K[x,y,z]/J$.

Example 2. 

Consider the ideal $I=\langle x^3-yz,y^3-xz,-x^{2}y+z^2,x^2{y}^2-z\rangle \subseteq K[x,y,z]$.

Using Maple, we compute the Gröbner basis of the ideal I in the quotient ring $K[x,y,z]/I$ as

$$G=[xz-z,yz-z,z^2-z,x^3-z,x^{2}y-z,y^3-z],$$

its monomial basis as $[1,x,y,z,x^2,yx,y^2,y^{2}x]$. Thus, the set of candidate elements is $T=\{t=x+ky+k^{2}z\mid 0⩽k⩽56\}$.

Implementing Algorithm 1 with Maple, the algorithm terminates at $k=56$, we find that for all candidate elements, the determinant of the matrix corresponding to their coefficients is zero. We conclude that the width of the ideal I is at least 2, meaning that I does not admit a cyclic basis.

6. Conclusions

This paper takes zero-dimensional ideals as the main research object, focusing on the problem of determining the existence of cyclic bases for their quotient rings. By combining the structural properties of quotient ring bases of zero-dimensional ideals when the ideal width is at most 1 with Gröbner basis theory, the study derives equivalent characterization conditions for the ideal width being at most 1. Further, using matrix theories, it designs an algorithm for determining the ideal width. Relying on the connection between ideal width being at most 1 and the existence of cyclic bases in quotient rings, the determination of the existence of cyclic bases in quotient rings is realized.

Compared with traditional methods, the method proposed in this paper designs an algorithm to judge the existence of a cyclic basis based on the relationship between the coefficients of the candidate elements and the width of the ideal. It reduces the scope of the candidate element set compared with the original method, provides a new approach for the cyclic basis judgment of the quotient ring of zero-dimensional ideals, and also offers new ideas for algorithm optimization in subsequent related fields.