Classification of Irreducible ${\mathbb{Z}}_{+}$-Modules of a ${\mathbb{Z}}_{+}$-Ring Using Matrix Equations

Abstract

This paper aims to investigate and categorize all inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules of a commutative unit ${\mathbb{Z}}_{+}$-ring $\mathcal{A}$, equipped with set {1, x, y, $xy$} satisfying $x^2=1,y^2=1$ as a ${\mathbb{Z}}_{+}$-basis by using matrix equations, which was part of a call for a Special Issue about matrix inequalities and equations by Symmetry. If the rank of the ${\mathbb{Z}}_{+}$-module $n\le 2$, we prove that there are finitely many inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules, respectively, one and three. However, if $n\ge 3$, there is no irreducible ${\mathbb{Z}}_{+}$-module.

1. Introduction

A tensor category is usually considered as a counterpart of groups and rings, and it is also important in algebraic representation theory. In 2000, Bakalov and Kirillov [1] gave an exposition of the relations between monoidal tensor categories and modular functors. In tensor categories, a ${\mathbb{Z}}_{+}$-ring is a quite significant invariant. In 1987, Lusztig [2] first introduced the ${\mathbb{Z}}_{+}$-ring. Afterwards, Etingof [3] proved in 1994 that irreducible ${\mathbb{Z}}_{+}$-representations of a Verlinde algebra which satisfies some conditions correspond to some special Dynkin diagrams. Davydov [4,5] showed the relation between the Drinfel’d algebras or Galois algebras and finite groups. The definitions and properties relevant to ${\mathbb{Z}}_{+}$-rings can be found in [3,6]. Particularly, Grothendieck rings [7,8,9,10,11] and Green rings of Hopf algebras [12,13,14,15,16,17] are both classic examples. Furthermore, Ostrik and others [3,6] introduced the terminology on ${\mathbb{Z}}_{+}$-modules over certain rings or algebras. If a ${\mathbb{Z}}_{+}$-module M only has a trivial ${\mathbb{Z}}_{+}$-submodule, it is called irreducible. Given a ring or algebra, it is important and meaningful to investigate and categorize all inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules over it.

In 2003, Ostrik [6] illustrated that given an arbitrary ${\mathbb{Z}}_{+}$-ring whose rank is finite, there are finitely many inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules. He stated that there is an upper bound for the rank of any irreducible ${\mathbb{Z}}_{+}$-module. Moreover, in general, the upper bound is quite large. Usually, it is quite difficult to estimate the size between the rank of a ${\mathbb{Z}}_{+}$-ring and that of an irreducible ${\mathbb{Z}}_{+}$-module. In this paper, we categorize all inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules of a commutative unit ${\mathbb{Z}}_{+}$-ring $\mathcal{A}$, equipped with set {1, x, y, $xy$} satisfying relations $x^2=1,y^2=1$ as a ${\mathbb{Z}}_{+}$-basis. In addition, this problem can be transformed to solve all the irreducible NIM (nonnegative integer matrix) solutions to matrix equations as follows:

$$\left\{\begin{array}{c}AB=BA,\\ A^2=E,\\ B^2=E,\end{array}\right.$$

(1)

where the matrix E is determined as a unit matrix through the paper. Hence, in what follows, we only need to determine all the irreducible NIM solutions to $\left(1\right)$. Particularly, Wang [18,19] made great contributions in matrix equations. In 2022, he derived the formula of a general solution to a class of equations named Sylvester type matrix equation of Hamilton quaternions and solvability conditions. Furthermore, he put forward a modified conjugate residual method to solve a class of equations named generalized coupled Sylvester tensor equations. In addition, Wang [20] investigated the minimum-norm least squares solution to a quaternion tensor system by the method of Moore–Penrose inverses in Symmetry.

In Section 3, we prove that there is no irreducible NIM solution to the matrix Equations $\left(1\right)$ when order $n\ge 5$, and there is no irreducible ${\mathbb{Z}}_{+}$-module of $\mathcal{A}$. In Section 4, when $n\le 4$, we first calculate all NIM solutions and then find the inequivalent irreducible ones. Furthermore, there are irreducible ${\mathbb{Z}}_{+}$-modules of $\mathcal{A}$ when and only when the rank of the ${\mathbb{Z}}_{+}$-module is no more than two. If $n=1$, there exists a unique irreducible NIM solution to $\left(1\right)$, which corresponds to a unique irreducible ${\mathbb{Z}}_{+}$-module of $\mathcal{A}$. If $n=2$, there are three inequivalent irreducible NIM solutions to the matrix equations $\left(1\right)$, which corresponds to three inequivalent irreducible ${\mathbb{Z}}_{+}$-modules of $\mathcal{A}$. If $n=3$ or $n=4$, there is no irreducible ${\mathbb{Z}}_{+}$-module of $\mathcal{A}$. In fact, this paper deals with semiring theory including semiring modules. This theory is closely related with tropical algebras theory. Particularly, Alexander Mikhalev [21,22] developed semigroup theory and semiring theory.

2. Preliminaries

Throughout this paper, all matrices are in ${\mathbb{M}}_n$($\mathbb{N}$), which is the set of matrices of order n over natural numbers, and the element of row i and column j of $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$ is denoted by $a_{ij}$. A is said to be positive if $a_{ij}>0$ for all i, j, and denoted by $A>0$. If $a_{ij}⩾0$ for all i, j, then A is said to be nonnegative, and denoted by $A⩾0$. Moreover, an n-order square matrix A is called a nonnegative integer matrix (NIM) if $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$. An n-order square matrix P is called a permutation matrix if every row and every column contains a single one and all the other elements are zero.

Definition 1

([23], Chapter XIII). Let A be an n-order matrix. Then, A is said to be reducible, if there is a permutation matrix P, satisfying $PA{P}^T=\left[\begin{array}{cc}{A}_1\hfill & O\hfill \\ A_2\hfill & A_3\hfill \end{array}\right]$, where $A_1$ is a k-order square matrix and $A_3$ is an $(n-k)$-order square matrix. Note that the submatrix O is a zero matrix and $PA{P}^T$ is a lower triangular block matrix.

Definition 2.

Let A and B be solutions to the system of matrix equations $\left(1\right)$. The solutions are said to be reducible, if A and B are simultaneously reducible, namely, there is an n-order permutation matrix P, satisfying

$$PA{P}^T=\left[\begin{array}{cc}{A}_1\hfill & O\hfill \\ A_2\hfill & A_3\hfill \end{array}\right],PB{P}^T=\left[\begin{array}{cc}{B}_1\hfill & O\hfill \\ B_2\hfill & B_3\hfill \end{array}\right],$$

where $A_1$ and $B_1$ are both square matrices with the same order and $A_3$ and $B_3$ are both square matrices with the same order. Otherwise, they are said to be irreducible.

From now on, let ${\mathbb{Z}}_{+}$ represent the set of nonnegative integers. In the following, we will present some basic definitions and properties relevant to a ${\mathbb{Z}}_{+}$-ring and a ${\mathbb{Z}}_{+}$-module [6].

Definition 3.

Let $\mathcal{A}$ be a ring and free as a $\mathbb{Z}$-module. A $\mathbb{Z}$-basis $\mathcal{B}={\left\{b_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ of $\mathcal{A}$ is called a ${\mathbb{Z}}_{+}$-basis if it satisfies $b_{\alpha }{b}_{\beta }={\sum }_{\omega \in \mathsf{\Omega }}{c}_{\alpha \beta }^{\omega }{b}_{\omega }$, and $c_{\alpha \beta }^{\omega }\in {\mathbb{Z}}_{+}$. A ${\mathbb{Z}}_{+}$-ring is a ring equipped with a special ${\mathbb{Z}}_{+}$-basis and identity one, which can be expressed by a nonnegative linear combination of it.

Definition 4.

(1) Assume $\mathcal{A}$ is a fixed ${\mathbb{Z}}_{+}$-ring equipped with a certain basis ${\left\{b_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$. An $\mathcal{A}$-module M is called a ${\mathbb{Z}}_{+}$-module if it is equipped with a special $\mathbb{Z}$-basis ${\left\{m_{\gamma }\right\}}_{\gamma \in \mathsf{\Gamma }}$ such that all the coefficients $t_{\omega \alpha }^{\gamma }$ defined by the $\mathbb{Z}$-basis relations $b_{\omega }{m}_{\alpha }={\sum }_{\gamma \in \mathsf{\Gamma }}{t}_{\omega \alpha }^{\gamma }{m}_{\gamma }$ are nonnegative.

(2) Assume $\mathcal{A}$ is a fixed ${\mathbb{Z}}_{+}$-ring equipped with a certain basis ${\left\{b_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ such that $b_{\alpha }{b}_{\beta }={\sum }_{\omega \in \mathsf{\Omega }}{c}_{\alpha \beta }^{\omega }{b}_{\omega }$. For a ${\mathbb{Z}}_{+}$-module M over $\mathcal{A}$, it assigns each $b_{\alpha }$ to a matrix $N_{\alpha }\in {\mathbb{M}}_n\left(\mathbb{N}\right)$ such that $N_{\alpha }{N}_{\beta }={\sum }_{\omega \in \mathsf{\Omega }}{c}_{\alpha \beta }^{\omega }{N}_{\omega }$, for all $\alpha ,\beta ,\omega \in \mathsf{\Omega }$. In particular, the identity of $\mathcal{A}$ is assigned to the unit matrix. It is also easy to see the rank of a ${\mathbb{Z}}_{+}$-module M equals the order of the matrix $N_{\omega }$.

Definition 5.

Take ${\mathbb{Z}}_{+}$-modules M, $M^{\prime }$ over $\mathcal{A}$ equipped with bases ${\left\{m_{\gamma }\right\}}_{\gamma \in \mathsf{\Gamma }}$ and ${\left\{m_{\beta }^{\prime }\right\}}_{\beta \in {\mathsf{\Gamma }}^{\prime }}$. They are said to be equivalent if there is a bijection $\psi :\mathsf{\Gamma }\to {\mathsf{\Gamma }}^{\prime }$ such that the induced $\mathbb{Z}$-linear map $\tilde{\psi }$ of $\mathbb{Z}$-modules M, $M^{\prime }$ defined by $\tilde{\psi }\left(m_{\gamma }\right)=m_{\psi \left(\gamma \right)}^{\prime }$ is an $\mathcal{A}$-isomorphism. Namely, ${\mathbb{Z}}_{+}$-modules M and $M^{\prime }$ of rank n are equivalent if and only if there is an n-order permutation matrix P such that $N_{\omega }^{\prime }=P{N}_{\omega }{P}^{-1}$, for all $\omega \in \mathsf{\Omega }$.

Definition 6.

(1) Let M be a ${\mathbb{Z}}_{+}$-module equipped with a basis ${\left\{m_{\gamma }\right\}}_{\gamma \in \mathsf{\Gamma }}$ and N be an $\mathcal{A}$-submodule of M. Then, N is called a ${\mathbb{Z}}_{+}$-submodule if it is spanned by ${\left\{m_{\gamma }\right\}}_{\gamma \in {\mathsf{\Gamma }}^{\prime }}$ as a subgroup of M, where ${\mathsf{\Gamma }}^{\prime }$ is a subset of Γ.

(2) A ${\mathbb{Z}}_{+}$-module M is said to be irreducible if it only has trivial ${\mathbb{Z}}_{+}$-submodule 0 or M. Otherwise, it is said to be irreducible. In fact, it is equivalent to the fact that all $N_{\omega }$ cannot be simultaneously transformed into lower triangular matrices, by Definition 2. Thus, it is called an irreducible NIM solution.

To prove our main theorems in the next sections, we will need the following statements.

Theorem 1

([23], Chapter XIII). Assume $A={\left(a_{ij}\right)}_{n\times n}$ is an n-order nonnegative real matrix. Then, A is irreducible if and only if ${(E+A)}^{n-1}>0$.

Theorem 2

([23], Chapter XIII). Assume $A={\left(a_{ij}\right)}_{n\times n}$ is an n-order nonnegative irreducible real matrix. Then, there is an eigenvalue λ of A such that λ is a real number and $\lambda >0$. Moreover, λ is a simple root such that λ$\ge \left|\alpha \right|$, where α is any eigenvalue of A. The eigenvalue λ is called a Perron–Frobenius eigenvalue.

Proposition 1

([24], Proposition 2.1). Let $A,B\in {\mathbb{M}}_n\left(\mathbb{N}\right)$ and $AB=E$, then A and B are both permutation matrices.

Corollary 1.

If $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$ and $A^k=E$, $k\ge 1$, then A is a permutation matrix.

Proof. 

Let $B=A^{k-1}$. Then, $AB=A^k=E$. From Proposition 1, A is a permutation matrix. □

3. Irreducible NIM Solutions for $\mathit{n}\ge \mathbf{5}$

Now, we infer that if the rank of ${\mathbb{Z}}_{+}$-ring $\mathcal{A}$, $n\ge 5$, equipped with a basis {1, x, y, $xy$} satisfying $x^2=1$, $y^2=1$, then there is no irreducible ${\mathbb{Z}}_{+}$-module M. This question is equivalent to proving that there is no irreducible NIM solution to the following equations:

$$\left\{\begin{array}{c}AB=BA,\\ A^2=E,\\ B^2=E,\end{array}\right.$$

(2)

where A and B∈${\mathbb{M}}_n\left(\mathbb{N}\right)$ are NIM solutions to the matrix equations $\left(2\right)$. Now, in the following, we let $b=E+A+B+AB$. Then, it is easy to see that $b\in {\mathbb{M}}_n\left(\mathbb{N}\right)$.

Theorem 3.

If A and B are irreducible NIM solutions, then b is irreducible.

Proof. 

Assume that b is reducible. Hence, we can get that there exists some permutation matrix P, satisfying

$$Pb{P}^T=P(E+A+B+AB)P^T=E+PA{P}^T+PB{P}^T+PAB{P}^T=\left[\begin{array}{cc}{b}_1\hfill & O\hfill \\ b_2\hfill & b_3\hfill \end{array}\right].$$

Since A, B and $AB$ are nonnegative integer matrices, we have

$$PA{P}^T=\left[\begin{array}{cc}{A}_1\hfill & O\hfill \\ A_2\hfill & A_3\hfill \end{array}\right],PB{P}^T=\left[\begin{array}{cc}{B}_1\hfill & O\hfill \\ B_2\hfill & B_3\hfill \end{array}\right],PAB{P}^T=\left[\begin{array}{cc}{C}_1\hfill & O\hfill \\ C_2\hfill & C_3\hfill \end{array}\right].$$

In this case, A and B are reducible solutions, which is a contradiction. Therefore, b is irreducible. □

From Theorem 3, we can get the following theorem.

Theorem 4.

If A and B are irreducible NIM solutions, then $b>0$.

Proof. 

From Theorem 3, we obtain that b is an irreducible nonnegative matrix. Thus, by Theorem 1, we can get ${(E+b)}^{n-1}>0$. According to the binomial expansion and the system of matrix equations $\left(2\right)$, we have

$${(E+b)}^{n-1}=m_{1}E+m_{2}A+m_{3}B+m_{4}AB>0,m_1,m_2,m_3,m_4\in {\mathbb{Z}}_{+}.$$

Now, we infer that $b=E+A+B+AB>0$. Assume that there is an element $b_{ij}$ of matrix b such that $b_{ij}=0$. Since A, B and $AB$ are nonnegative integer matrices, the elements of row i and column j for each component $E,A,B$ and $AB$ are all 0, which is a contradiction. Therefore, we can get $b=E+A+B+AB>0$. □

Furthermore, we can deduce the following theorem.

Theorem 5.

If $b>0$, then b is irreducible and $tr\left(b\right)=4$.

Proof. 

From Theorem 1, b is irreducible. According to the system of matrix equations $\left(2\right)$, we have $b^2=4b$. Assume that $\lambda$ is an eigenvalue of b. Then, we have ${\lambda }^2=4\lambda$, that is $\lambda =4\mathrm{or}0$. Moreover, we claim that 4 is an eigenvalue of b. Otherwise, the trace of b, $tr\left(b\right)=0$, which contradicts $b>0$ and $tr\left(b\right)>0$. Considering that b is irreducible and $b>0$, by Theorem 2, we infer that 4 is the unique nonzero simple root of b. Therefore, we can get $tr\left(b\right)=4$. □

In what follows, we give our prime theorem in this chapter.

Theorem 6.

Let $A,B\in {\mathbb{M}}_n\left(\mathbb{N}\right)$ be solutions to the matrix equations $\left(2\right)$ and $n\ge 5$. Then, A and B are reducible solutions.

Proof. 

Assume that there are irreducible NIM solutions A and B to the matrix equations $\left(2\right)$. By Theorem 4, we get $b>0$. Therefore, $tr\left(b\right)\ge tr\left(E\right)=n\ge 5$, which contradicts the fact that $tr\left(b\right)=4$ by Theorem 5. Hence, the NIM solutions are always reducible. □

4. Irreducible NIM Solutions for $\mathit{n}\le \mathbf{4}$

We have already illustrated that there is no irreducible NIM solution to the matrix equations $\left(2\right)$ for $n\ge 5$. In the following, we describe all the inequivalent and irreducible NIM solutions to $\left(2\right)$, for $n\le 4$.

• If $n=1$, there is a unique irreducible NIM solution:

  $$A=1,B=1.$$
• If $n=2$, according to the fact that $A^2$ = E and $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$, we have that A or B is one of the two matrices as follows:

  $$\left(1\right)\left[\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]\left(2\right)\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right].$$

Since there is a bijection between permutation matrices $P_2$ and elements of permutation group $S_2$, we can denote the matrices above by $\left(1\right)$ and $\left(12\right)$, respectively. We also know that B is one of the two matrices. By the commutativity of A and B, there are four possible NIM solutions to the matrix equations $\left(2\right)$. Furthermore, there are a total of three irreducible NIM solutions to the matrix equations $\left(2\right)$, except that both A and B are unit matrices:

• Case(1.1):$A_1=\left(1\right)$, $B_1=\left(12\right)$.
• Case(1.2):$A_2=\left(12\right)$, $B_2=\left(1\right)$ or $B_3=\left(12\right)$.

• If $n=3$, since $A^2=E$ and $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$, we infer that A is one of the following four matrices:
  $\left(1\right)\left[\begin{array}{ccc}0\hfill & 1\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right]\left(2\right)\left[\begin{array}{ccc}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 1\hfill & 0\hfill \end{array}\right]\left(3\right)\left[\begin{array}{ccc}0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill \end{array}\right]\left(4\right)\left[\begin{array}{ccc}1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill \end{array}\right].$

Since there is a bijection between permutation matrices $P_3$ and elements of permutation group $S_3$, we can denote the matrices above by $\left(12\right)$, $\left(23\right)$, $\left(13\right)$ and $\left(1\right)$, respectively. Thus, B is one of the four matrices. According to the commutativity of A and B, we get the following NIM solutions to the matrix equations $\left(2\right)$:

• Case(2.1):$A_1=\left(12\right)$, then $B_1=\left(1\right)$ or $B_2=\left(12\right)$.
• Case(2.2):$A_2=\left(23\right)$, then $B_3=\left(1\right)$ or $B_4=\left(23\right)$.
• Case(2.3):$A_3=\left(13\right)$, then $B_5=\left(1\right)$ or $B_6=\left(13\right)$.
• Case(2.4):$A_4=\left(1\right)$, then $B_7=\left(1\right)$ or $B_8=\left(12\right)$ or $B_9=\left(13\right)$ or $B_{10}=\left(23\right)$.

Furthermore, we find that all the NIM solutions above are reducible. Hence, if $n=3$, then there is no irreducible NIM solution to the matrix equations $\left(2\right)$.

• If $n=4$, since $A^2=E$ and $A\in {\mathbb{M}}_n\left(\mathbb{N}\right)$, we get that A is one of the following ten matrices:
  (1)$\left[\begin{array}{cccc}1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]$ (2)$\left[\begin{array}{cccc}1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \end{array}\right]$ (3)$\left[\begin{array}{cccc}0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]$ (4)$\left[\begin{array}{cccc}0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right]$
  (5)$\left[\begin{array}{cccc}1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \end{array}\right]$ (6)$\left[\begin{array}{cccc}0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \end{array}\right]$ (7)$\left[\begin{array}{cccc}0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \end{array}\right]$ (8)$\left[\begin{array}{cccc}0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \end{array}\right]$
  (9)$\left[\begin{array}{cccc}0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \\ 1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \end{array}\right]$(10)$\left[\begin{array}{cccc}1\hfill & 0\hfill & 0\hfill & 0\hfill \\ 0\hfill & 1\hfill & 0\hfill & 0\hfill \\ 0\hfill & 0\hfill & 1\hfill & 0\hfill \\ 0\hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right].$

Since there is a bijection between permutation matrices $P_4$ and elements of permutation group $S_4$, we can denote the matrices above by $\left(23\right)$, $\left(24\right)$, $\left(13\right)$, $\left(12\right)$, $\left(34\right)$, $\left(14\right)\left(23\right)$, $\left(14\right)$, $\left(12\right)\left(34\right)$, $\left(13\right)\left(24\right)$ and $\left(1\right)$, respectively. It is easy to see that B is one of the ten matrices. By the commutativity of A and B, we get the following NIM solutions to the matrix equations $\left(2\right)$:

• Case(3.1): If $A_1=\left(23\right)$, then $B_1=\left(1\right)$ or $B_2=\left(14\right)$ or $B_3=\left(23\right)$ or $B_4=\left(14\right)\left(23\right)$.
• Case(3.2): If $A_2=\left(24\right)$, then $B_5=\left(1\right)$ or $B_6=\left(24\right)$ or $B_7=\left(13\right)$ or $B_8=\left(13\right)\left(24\right)$.
• Case(3.3): If $A_3=\left(13\right)$, then $B_9=\left(1\right)$ or $B_{10}=\left(13\right)$ or $B_{11}=\left(24\right)$ or $B_{12}=\left(13\right)\left(24\right)$.
• Case(3.4): If $A_4=\left(12\right)$, then $B_{13}=\left(1\right)$ or $B_{14}=\left(34\right)$ or $B_{15}=\left(12\right)$ or $B_{16}=\left(12\right)\left(34\right)$.
• Case(3.5): If $A_5=\left(34\right)$, then $B_{17}=\left(1\right)$ or $B_{18}=\left(34\right)$ or $B_{19}=\left(12\right)$ or $B_{20}=\left(12\right)\left(34\right)$.
• Case(3.6): If $A_6=\left(14\right)\left(23\right)$, then $B_{21}=\left(1\right)$ or $B_{22}=\left(14\right)\left(23\right)$ or $B_{23}=\left(14\right)$ or $B_{24}=\left(23\right)$.
• Case(3.7): If $A_7=\left(14\right)$, then $B_{25}=\left(1\right)$ or $B_{26}=\left(14\right)$ or $B_{27}=\left(23\right)$ or $B_{28}=\left(14\right)\left(23\right)$.
• Case(3.8): If $A_8=\left(12\right)\left(34\right)$, then $B_{29}=\left(1\right)$ or $B_{30}=\left(12\right)\left(34\right)$ or $B_{31}=\left(12\right)$ or $B_{32}=\left(34\right)$.
• Case(3.9): If $A_9=\left(13\right)\left(24\right)$, then $B_{33}=\left(1\right)$ or $B_{34}=\left(13\right)\left(24\right)$ or $B_{35}=\left(13\right)$ or $B_{36}=\left(24\right)$.
• Case(3.10): If $A_{10}=\left(1\right)$, then $B_{37}=\left(23\right)$ or $B_{38}=\left(24\right)$ or $B_{39}=\left(13\right)$ or $B_{40}=\left(12\right)$ or $B_{41}=\left(34\right)$ or $B_{42}=\left(14\right)\left(23\right)$ or $B_{43}=\left(14\right)$ or $B_{44}=\left(12\right)\left(34\right)$ or $B_{45}=\left(13\right)\left(24\right)$ or $B_{46}=\left(1\right)$.

Furthermore, we obtain that all the NIM solutions above are reducible. Hence, if $n=4$, there is no irreducible NIM solution to the matrix equations $\left(2\right)$.

According to the discussion above, we give all the irreducible NIM solutions to the matrix equations $\left(2\right)$, where $n\le 4$, and get the following theorem.

Theorem 7.

All the irreducible NIM solutions to the matrix equations $\left(2\right)$ are as follows:

Now, according to

Table 1. All the irreducible NIM solutions to $\left(2\right)$.

All the Irreducible NIM Solutions to (2)
n	A	B
1	1	1
2	$\left[\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$	$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$
	$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$	$\left[\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right],$$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$
≥3	None	None

, we successfully categorize all the irreducible ${\mathbb{Z}}_{+}$-modules of $\mathcal{A}$ under the equivalence by Definition 5, and in conclusion we have the main theorem below.

Theorem 8.

All the inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules of ${\mathbb{Z}}_{+}$-ring $\mathcal{A}$ are as follows (see

Table 2. All the inequivalent and irreducible ${\mathbb{Z}}_{+}$-modules of ${\mathbb{Z}}_{+}$-ring $\mathcal{A}$.

All the Inequivalent and Irreducible ${\mathbb{Z}}_{+}$-Modules of $\mathcal{A}$
n	A	B
1	1	1
2	$\left[\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right]$	$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$
	$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$	$\left[\begin{array}{cc}1\hfill & 0\hfill \\ 0\hfill & 1\hfill \end{array}\right],$$\left[\begin{array}{cc}0\hfill & 1\hfill \\ 1\hfill & 0\hfill \end{array}\right]$
≥3	None	None

):