Ideals in Bipolar Quantum Linear Algebra

Abstract

Since bipolar quantum linear algebra (BQLA), under two operations–-addition and multiplication—demonstrates the properties of semirings, and since ideals play an important role in abstract algebra, our results are compelling for the ideals of a semiring. In this article, we investigate the characteristics of ideals, principal ideals, prime ideals, maximal ideals, and the smallest ideal containing any nonempty subset. By applying elementary real analysis, particularly the infimum, our main result is stated as follows: for any closed set I in BQLA, I is a nontrivial proper ideal if and only if there exists $c\in (0,1]$ such that $I=\left\{(-x,y)\in {\mathbb{R}}^2|cx\le y\le {\displaystyle \frac{x}{c}}\mathrm{and}x,y\ge 0\right\}$. This shows that its shape has to be symmetric with the graph $y=-x$.

1. Introduction

In ring theory, ideals play a significant role in studying various properties and structures of rings, such as quotient rings and factor structures. In essence, the study and exploration of ideals in semirings not only reveal the intricate details of these algebraic structures but also provide a versatile framework for theoretical advancements and practical applications across diverse fields of mathematics and beyond. Previous research has studied the use of knowledge of ideals. In 2014, Y. Katsov, T. G. Nam, and J. Zumbrägel [1] investigated various classes of semirings and complete semirings, focusing on the property of being ideal-simple, congruence-simple, or both. In 2018, P. Nasehpour [2] examined the prime, primary, and maximal ideals of commutative semirings. The localization and primary decomposition of ideals in commutative semirings were also studied. In 2019, I. Chajda and H. Länger [3] studied complements of annihilator ideals and constructed commutative semirings with a complemented lattice of ideals. In 2023, M. Korbelář [4] studied the idempotent and torsion properties with additively divisible elements in a finitely generated commutative semiring with an identity. Additionally, J. Goswami and L. Boro [5] showed a weakly nilpotent graph on a commutative semiring with an identity. Despite decades of research, the properties of ideals continue to be discussed and applied as the foundation for in-depth study in different fields of mathematics.

In 2011, W.R. Zhang proposed the idea of bipolar quantum linear algebra (BQLA). BQLA can be applied to various fields such as bipolar fuzzy sets, bipolar disorder, bipolar cognitive mapping, and the causal theory of YinYang bipolar atoms, which includes supper symmetrical quantum cosmology. Moreover, in Chapter 8 of [6], BQLA was an important tool for creating many results; for example, the symmetry law (or elementary energy equilibrium), the energy transfer equilibrium law [7], and the law of energy symmetry (or YinYang-n-element system nonequilibrium strengthening law) [8]. Some recent articles related to the bipolar concepts, for example, “Ground-0 Axioms vs. First Principles and Second Law: From the Geometry of Light and Logic of Photon to Mind–Light–Matter Unity-AI&QI” [9] and “Bipolar Fuzzy Supra Topology via (Q-) Neighborhood and its Application in Data Mining Process” [10]. Even though BQLA has been widely applied in many fields, we found that BQLA is hardly developed in the field of ideals. So, this is one of our aspirations to study the system of BQLA in the theoretical sense of abstract algebra that may be applied for further applications.

This article aims to study the ideal characteristic of I, where I is either an open or closed set in a bipolar set B with respect to the usual topology on ${\mathbb{R}}^2$, and is divided into three parts as follows:

• We give the necessary and sufficient conditions to analyze the ideal I (see Theorem 3 and Corollary 2), and the consequences are all possible ideals I that can be expressed in the symmetric region with the graph $y=-x$.
• All principal ideals, prime ideals, and maximal ideals are identified in Theorem 4, Theorem 5, and Theorem 6, respectively.
• If $I\ne \left\{\right(0,0\left)\right\}$ is a closed subset in $\left\{(-a,b)\right|a,b\in {\mathbb{R}}^{+}\}$, then we also deliver the conditions to classify the ideal generated by I into three forms: $\left\{(-a,b)\right|a,b\in {\mathbb{R}}^{+}\}\cup \left\{(0,0)\right\}$, $\left\{\right(-x,x)\in B\}$, or $\left\{(-x,y)\in B|cx\le y\le {\displaystyle \frac{x}{c}}\right\}$ for some $c\in (0,1]$ (see Theorem 9).

Throughout the proofs in this article, we try to limit all tools to those easily accessible to students pursuing an undergraduate degree. We use basic knowledge of elementary mathematical analysis, especially open sets, closed sets, and the usual topology on ${\mathbb{R}}^2$. In addition, we apply the concept of an infimum to claim the existence of a constant $c\in (0,1]$ for representing the ideal $I=\left\{(-x,y)\in {\mathbb{R}}^2|cx\le y\le {\displaystyle \frac{x}{c}}\mathrm{and}x,y\ge 0\right\}$.

The outline of this paper is as follows. We provide the definitions and terminology required for this research in Section 2. Our significant findings concerning the characteristics of ideals, principal ideals, prime ideals, maximal ideals, and the smallest ideal containing any nonempty subset are demonstrated in Section 3. This section is divided into two parts: first, the proofs utilize basic mathematical methods and apply elementary real analysis with a focus on the infimum; and second, we incorporate a suggestion from a reviewer, which was to simplify our proofs by applying an isomorphism of semirings, as presented in Section 3.2. In the last section, Section 4, we summarize the overall perspective of this article.

2. Material and Methods

For simplicity, below, we present the terminology [6,11,12] used in this article.

Definition 1.

The set of all bipolar agents is the bipolar set

$$B=\left\{(-a,b)\right|a,b\in {\mathbb{R}}_0^{+}\}.$$

The interior of B is $B^{*}=\left\{(-a,b)\right|a,b\in {\mathbb{R}}^{+}\}$.

Definition 2

(Addition of the bipolar set). Let $(-a,b),(-c,d)\in B$. Then,

$$(-a,b)+(-c,d)=(-a-c,b+d)$$

with $(0,0)$ as the identity (see Figure 1). The addition has commutative and associative properties.

Definition 3

(Multiplication of the bipolar set). Let $(-a,b),(-c,d)\in B$. Then,

$$(-a,b)(-c,d)=(-ad-bc,ac+bd)$$

with $(0,1)$ as the identity (see Figure 2). The multiplication has commutative and associative properties. Moreover, the multiplication has distributive properties over the addition.

Definition 4.

Let R be a nonempty set. Then, $(R,+,·)$ is a semiring if the following conditions hold:

1. 

$(R,+)$ is a commutative semigroup;

2. 

$(R,·)$ is a semigroup;

3. 

Left and right distributivity hold.

In particular, a semiring $(R,+,·)$ is said to be commutative if $(R,·)$ is commutative. If a semigroup $(R,·)$ has a neutral element, we call this element the identity of the semiring R. If a semigroup $(R,+)$ has a neutral element, we call this element the zero of the semiring R.

As a result, $(B,+,·)$ forms a commutative semiring with identity $(0,1)$ and zero $(0,0)$, and $(B^{*},+,·)$ forms a commutative semiring.

Definition 5.

A nonempty subset I of a semiring R is said to be a left (resp. right) ideal of R if the following conditions hold:

1. 

$a+b\in I$ for all $a,b\in I$;

2. 

$ra$ (resp. $ar$) $\in I$ for any $a\in I$ and $r\in R$.

If I is a left and a right ideal of a semiring R, it is called a two-sided ideal, or simply, an ideal of R. In addition, if R is commutative, then the concepts of left ideals, right ideals, and two-sided ideals clearly coincide.

Definition 6.

A proper ideal P of a semiring R is prime if $ab\in P$ implies that $a\in P$ or $b\in P$ for all elements a and b of R.

Definition 7.

A proper ideal M of a semiring R is maximal if $M\subseteq I\subseteq R$ for any ideal I of R implies that either $I=M$ or $I=R$.

Definition 8.

For each nonempty subset X of a semiring R, ${\left(X\right)}_i$ is the smallest ideal of R containing X, called the ideal of R generated by X.

Definition 9.

An ideal I of a semiring R is principal if there exists an element a of R such that $I={\left(\left\{a\right\}\right)}_i$, or simply, $I={\left(a\right)}_i$.

3. Main Results

3.1. Proofs Utilizing Foundational Mathematical Techniques

First, we introduce the notations and some properties that play a significant role in this section.

Let $s,t\in {\mathbb{R}}^{+}$ be such that $s\ge t$. Then, we define (see Figure 3)

$$\overline{I_{(-s,t)}}=\left\{(-x,y)\in B|{\displaystyle \frac{t}{s}}x\le y\le {\displaystyle \frac{s}{t}}x\right\}$$

and

$$I_{(-s,t)}=\left\{(-x,y)\in B|{\displaystyle \frac{t}{s}}x<y<{\displaystyle \frac{s}{t}}x\right\}.$$

Actually, $\overline{I_{(-1,1)}}=\left\{(-x,x)\right|x\ge 0\}$. Moreover, we let the notations $\overline{I_{(0,b)}}=\overline{I_{(-a,0)}}=B$ and $I_{(0,b)}=I_{(-a,0)}=B^{*}$ for any $a,b\in {\mathbb{R}}^{+}$.

Proposition 1. 

Let $s,t\in {\mathbb{R}}^{+}$. Then, $\overline{I_{(-s,t)}}$ and $I_{(-s,t)}$ are closed under addition.

Proof. 

We only show the case $\overline{I_{(-s,t)}}$. Let $(-a,b),(-c,d)\in \overline{I_{(-s,t)}}$. Since ${\displaystyle \frac{b}{a}}\le {\displaystyle \frac{s}{t}}$ and ${\displaystyle \frac{d}{c}}\le {\displaystyle \frac{s}{t}}$, we obtain $bt\le as$ and $dt\le cs$. Then, $t(b+d)=bt+dt\le as+cs=s(a+c)$. Thus, ${\displaystyle \frac{b+d}{a+c}}\le {\displaystyle \frac{s}{t}}$. Similarly, it is easy to see that ${\displaystyle {\displaystyle \frac{b+d}{a+c}}\ge {\displaystyle \frac{t}{s}}}$. □

Theorem 1.

Let $s,t\in {\mathbb{R}}^{+}$. Then, $\overline{I_{(-s,t)}}$ is an ideal of a semiring B.

Proof. 

By Proposition 1, $\overline{I_{(-s,t)}}$ is closed under addition. Next, let $(-c,d)\in \overline{I_{(-s,t)}}$ and $(-x,y)\in B$. Then, $(-a,b):=(-c,d)(-x,y)$. So, $a=cy+dx$ and $b=cx+dy$. Consider ${\displaystyle {\displaystyle \frac{b}{a}}={\displaystyle \frac{\frac{c}{d}x+y}{\frac{c}{d}y+x}}}$. Since $(-c,d)\in \overline{I_{(-s,t)}}$, we have $(-d,c)\in \overline{I_{(-s,t)}}$ such that ${\displaystyle {\displaystyle \frac{t}{s}}\le {\displaystyle \frac{d}{c}}\le {\displaystyle \frac{s}{t}}}$ and ${\displaystyle {\displaystyle \frac{t}{s}}\le {\displaystyle \frac{c}{d}}\le {\displaystyle \frac{s}{t}}}$. If ${\displaystyle {\displaystyle \frac{d}{c}}\ge 1}$, then ${\displaystyle {\displaystyle \frac{b}{a}}<{\displaystyle \frac{d}{c}}\le {\displaystyle \frac{s}{t}}}$, and if ${\displaystyle {\displaystyle \frac{d}{c}}<1}$, then ${\displaystyle {\displaystyle \frac{b}{a}}<{\displaystyle \frac{c}{d}}\le {\displaystyle \frac{s}{t}}}$. So, we have ${\displaystyle {\displaystyle \frac{b}{a}}\le {\displaystyle \frac{s}{t}}}$. Similarly, if ${\displaystyle {\displaystyle \frac{c}{d}}\ge 1}$, then ${\displaystyle {\displaystyle \frac{b}{a}}={\displaystyle \frac{\frac{c}{d}x+y}{\frac{c}{d}y+x}}>{\displaystyle \frac{x+y}{\frac{c}{d}y+\frac{c}{d}x}}={\displaystyle \frac{d}{c}}\ge {\displaystyle \frac{t}{s}}}$, and if ${\displaystyle \frac{c}{d}}<1$, then ${\displaystyle \frac{b}{a}}={\displaystyle \frac{\frac{c}{d}x+y}{\frac{c}{d}y+x}}>{\displaystyle \frac{\frac{c}{d}x+\frac{c}{d}y}{x+y}}={\displaystyle \frac{c}{d}}\ge {\displaystyle \frac{t}{s}}$. Thus, ${\displaystyle \frac{b}{a}}\le {\displaystyle \frac{t}{s}}$. As a result, $(-a,b)\in \overline{I_{(-s,t)}}$. We conclude that $\overline{I_{(-s,t)}}$ is an ideal of a semiring B. □

Theorem 2.

Let $s,t\in {\mathbb{R}}^{+}$. Then, $I_{(-s,t)}$ is an ideal of a semiring $B^{*}$.

Proof. 

Applying Proposition 1 and a similar argument as Theorem 1. □

Corollary 1. 

Let $s,t\in {\mathbb{R}}^{+}$. Then, $I_{(-s,t)}\cup \left\{(0,0)\right\}$ is an ideal of a semiring B.

Now, we know that the following sets are ideals of the semiring B: $\left\{\right(0,0\left)\right\}$, B, $B^{*}\cup \left\{(0,0)\right\}$, $I_{(-s,t)}\cup \left\{(0,0)\right\}$, $\overline{I_{(-1,1)}}$, and $\overline{I_{(-s,t)}}$, where $s>t$ (see Figure 4 and Figure 5). In addition, the sets $B^{*}$, $I_{(-s,t)}$, $\overline{I_{(-1,1)}}\setminus \left\{(0,0)\right\}$, and $\overline{I_{(-s,t)}}\setminus \left\{(0,0)\right\}$, where $s>t$ are ideals of the semiring $B^{*}$ (see Figure 6). Moreover,

Table 1. Open and closed sets in B.

Ideals in a Semiring B	Open	Closed
$\left\{\right(0,0\left)\right\}$	✗	✓
B	✗	✓
$B^{*}\cup \left\{(0,0)\right\}$	✗	✗
$\overline{I_{(-s,t)}}$	✗	✓
$\overline{I_{(-1,1)}}$	✗	✓
$I_{(-s,t)}\cup \left\{(0,0)\right\}$	✗	✗

and

Table 2. Open and closed sets in $B^{*}$.

Ideals in a Semiring ${\mathit{B}}^{*}$	Open	Closed
$B^{*}$	✓	✗
$I_{(-s,t)}$	✓	✗
$\overline{I_{(-1,1)}}\setminus \left\{(0,0)\right\}$	✗	✓
$\overline{I_{(-s,t)}}\setminus \left\{(0,0)\right\}$	✗	✓

present the relatively open (closed) sets in B and $B^{*}$, respectively.

This motivates us to consider other forms of ideals. So, the remainder of this section is devoted to answering this query.

Theorem 3.

Let I be a closed set of a semiring B. Then, I is a nontrivial proper ideal of B if and only if there exists $0<c\le 1$ such that $I=\overline{I_{(-1,c)}}$.

Proof. 

We only prove the sufficient part. Since I is a closed set and a nontrivial ideal of B, we know that I is not $\left\{\right(0,0\left)\right\}$, $I_{(-s,t)}\cup \left\{(0,0)\right\}$, $B^{*}\cup \left\{(0,0)\right\}$, or B. We assume that there exists $(-x_0,y_0)\in I$ such that $x_0$ and $y_0$ are positive. If no such point $(-x_0,y_0)$ exists where $x_0\ne y_0$, then I must be $\left\{\right(-x,x\left)\right|x\ge 0\}$, that is, $I=\overline{I_{(-1,1)}}$. Now, we have $(-x_0,y_0)\in I$ such that $x_0\ne y_0$, and both of them are positive. Let $C=\left\{{\displaystyle \frac{y}{x}}|(-x,y)\in I\right\}$. Then, $0<c=infC<1$ because $(-x_0,y_0)\in I$, $(-y_0,x_0)=(-1,0)(-x_0,y_0)\in I$ and $min\left\{{\displaystyle \frac{x_0}{y_0}},{\displaystyle \frac{y_0}{x_0}}\right\}<1$. To show that $I=\overline{I_{(-1,c)}}$, we need more notations and rely on the following results: since the ideal $I\ne B$, there exists $(-a,b)\in B^{*}\setminus I$ such that $a\ne b$. If $a=b$, it would contradict the fact that I is an ideal. Specifically,

$$(-a,a)=\left(0,{\displaystyle \frac{a}{\alpha +\beta }}\right)\left(-(\alpha +\beta ),\alpha +\beta \right)\in I\mathrm{for}\mathrm{some}(-\alpha ,\beta )\in I.$$

Let $M=min\left\{{\displaystyle \frac{a}{b}},{\displaystyle \frac{b}{a}}\right\}<1$ and $m>0$ be such that $m\le M$. From now on, we consider only the case $a>b$. The case $a<b$ can be proven in a similar manner. Since $a>b$, we obtain ${\displaystyle M=\frac{b}{a}}$. Define $L_m=\{(-x,y)\in B^{*}|y=(-m)(-x)\}$. We claim the following:

• $L_m\cap I=\varnothing$ for all $0<m\le M$;
• $A\cap I=\varnothing$, where $A=\left\{(-x,y)\in B^{*}|M>{\displaystyle \frac{y}{x}}\right\}$;
• M is a lower bound of C.

Suppose that we have these claims; we are ready to show that $I=\overline{I_{(-1,c)}}$. Let $(-x,y)\in I$. Then, ${\displaystyle \frac{y}{x}}\in C$, and we have $c\le {\displaystyle \frac{y}{x}}$. If ${\displaystyle \frac{y}{x}}\le 1$, then $c\le {\displaystyle \frac{y}{x}}\le 1\le {\displaystyle \frac{1}{c}}$. Therefore, $(-x,y)\in \overline{I_{(-1,c)}}$. If ${\displaystyle \frac{y}{x}}>1$, then $0<{\displaystyle \frac{x}{y}}<1$. Since I is an ideal and $(-x,y)\in I$, we obtain $(-y,x)\in I$. So, ${\displaystyle \frac{x}{y}}\in C$, and then $c\le {\displaystyle \frac{x}{y}}<1\le {\displaystyle \frac{1}{c}}$. Thus, $(-y,x)\in \overline{I_{(-1,c)}}$. Since $\overline{I_{(-1,c)}}$ is an ideal, we obtain $(-x,y)\in \overline{I_{(-1,c)}}$. Using the contrapositive, we assume that $(-x,y)\notin I$. Applying a similar argument as for the point $(-a,b)\notin I$, we conclude that $M^{\prime }=min\left\{{\displaystyle \frac{x}{y}},{\displaystyle \frac{y}{x}}\right\}$ is a lower bound for C. Since I is a closed set, we have $c=minC\in C$, and then ${\displaystyle \frac{y}{x}}<c$ or ${\displaystyle \frac{y}{x}}>{\displaystyle \frac{1}{c}}$. So, $(-x,y)\notin \overline{I_{(-1,c)}}$. We can conclude that $\overline{I_{(-1,c)}}\subseteq I$. Now, we proceed to prove our claims as follows:

• For $m=M$, suppose $L_M\cap I\ne \varnothing$. We let $(-e,f)\in L_M\cap I$, and then $b=f\left({\displaystyle \frac{a}{e}}\right)$. Since $(-e,f)\in I$ and I is an ideal, we have $(-a,b)=\left(-e\left({\displaystyle \frac{a}{e}}\right),f\left({\displaystyle \frac{a}{e}}\right)\right)=\left(0,{\displaystyle \frac{a}{e}}\right)(-e,f)\in BI\subseteq I$, which contradicts $(-a,b)\notin I$. For $m<M$, suppose $(-e,f)\in L_m\cap I$. Since I is an ideal, $(-e,f)\in I$, and $m={\displaystyle \frac{f}{e}}$, we have $(-1,m)=\left(0,{\displaystyle \frac{1}{e}}\right)(-e,f)\in I$. If we can choose $(-s,t)\in B^{*}$ such that $(-1,M)=(-s,t)(-1,m)$, then $(-1,M)\in I$, but $(-1,M)\in L_M$. This contradicts $L_M\cap I=\varnothing$. We select s and t to finish the proof. Choose $s={\displaystyle \frac{M-m}{1-m^2}}$ and $t=1-ms=m\left({\displaystyle \frac{1}{m}}-s\right)$. Then, $s>0$ and $1-M>0>m(m-1)=m^2-m$. Thus, $1-m^2>M-m>0$, and then ${\displaystyle \frac{1}{m}}>1>{\displaystyle \frac{M-m}{1-m^2}}=s$. So, $t>0$. Hence, $(-s,t)\in B^{*}$ and

  $$\begin{array}{cc}\hfill (-s,t)(-1,m)& =(-sm-t,s+mt)\hfill \\ \hfill & =\left(-sm-1+ms,{\displaystyle \frac{M-m}{1-m^2}}+m-m^2\left({\displaystyle \frac{M-m}{1-m^2}}\right)\right)\hfill \\ \hfill & =\left(-1,{\displaystyle \frac{M-m+m-m^3-m^{2}M+m^3}{1-m^2}}\right)=(-1,M).\hfill \end{array}$$
• It is easy to see that $A\cap I=\varnothing$ since ${\displaystyle A=\bigcup _{0<m<M}{L}_m}$ and $L_m\cap I=\varnothing$ for all $0<m<M$.
• Suppose that M is not a lower bound of C. Then, there exists $(-x,y)\in I$ such that ${\displaystyle \frac{y}{x}}<M$. Consequently, $(-x,y)\in A$. Hence, $(-x,y)\in A\cap I$, which is a contradiction.

□

Corollary 2. 

Let $I\setminus \left\{\right(0,0\left)\right\}$ be an open set of a semiring B. Then, I is a nontrivial ideal of B if and only if there exists $0<c<1$ such that $I=I_{(-1,c)}\cup \left\{(0,0)\right\}$ or $I=B^{*}\cup \left\{(0,0)\right\}$.

Proof. 

In Theorem 3, by changing the set $C=\left\{{\displaystyle \frac{y}{x}}|(-x,y)\in I\right\}$ to a new set $C=\left\{{\displaystyle \frac{y}{x}}|(-x,y)\mathrm{is}\mathrm{any}\mathrm{element}\mathrm{in}\mathrm{the}\mathrm{closure}\mathrm{of}I\mathrm{and}(-x,y)\ne (0,0)\right\}$ and applying the same process in the proof of Theorem 3. □

Remark 1.

Let $a,b,c,d\in {\mathbb{R}}^{+}$. Then,

1. 

$\overline{I_{(-1,1)}}\subseteq \overline{I_{(-a,b)}}$.

2. 

$\overline{I_{(-1,1)}}=\overline{I_{(-a,a)}}$.

3. 

$\overline{I_{(-a,b)}}=\overline{I_{(-c,d)}}$ if ${\displaystyle \frac{b}{a}=\frac{d}{c}}$.

4. 

$\overline{I_{(-a,b)}}=\overline{I_{(-a,b)·(-c,d)}}$ if ${\displaystyle \frac{b}{a}=\frac{d}{c}}$.

5. 

If $0<a<b<1$, then $\overline{I_{(-1,b)}}\subseteq \overline{I_{(-1,a)}}$.

The principal ideals, prime ideals, and maximum ideals in a semiring B are discussed in the following theorems.

Theorem 4.

Let $s,t\in {\mathbb{R}}^{+}$. Then, $\overline{I_{(-s,t)}}$ is a principal ideal of a semiring B. Moreover, every nontrivial proper ideal of B, being a closed set, is a principal ideal.

Proof. 

We show that $\overline{I_{(-s,t)}}={\left((-s,t)\right)}_i$. If $s=t$, clearly $\overline{I_{(-1,1)}}={\left((-1,1)\right)}_i$. Suppose $t<s$. Since ${\left((-s,t)\right)}_i=\left\{(-x,y)(-s,t)\right|(-x,y)\in B^{*}\},$$(-s,t)\in \overline{I_{(-s,t)}}$, and $\overline{I_{(-s,t)}}$ is an ideal of a semiring B, it follows that ${\left((-s,t)\right)}_i\subseteq \overline{I_{(-s,t)}}$. Conversely, let $(-a,b)\in \overline{I_{(-s,t)}}$. Then,

$${\displaystyle \frac{t}{s}}a\le b\le {\displaystyle \frac{s}{t}}a.$$

(1)

Choose ${\displaystyle m=\frac{as-bt}{s^2-t^2}}$ and ${\displaystyle n=\frac{at-bs}{t^2-s^2}}$. Then, $m>0$ because $t<s$ and $bt<as$ by Equation (1). Similarly, we know that $n>0$ because $t<s$ and $at<bs$. Moreover, $(-t,s)=(-1,0)(s-,t)\in {\left((-s,t)\right)}_i$ and $(-s,t)\in {\left((-s,t)\right)}_i$. Thus, we obtain

$$(-a,b)=(0,m)(-s,t)+(0,n)(-t,s)\in {\left((-s,t)\right)}_i.$$

By Theorem 3, we conclude that every nontrivial proper ideal of B, being a closed set, is $\overline{I_{(-s,t)}}=\overline{I_{(-1,{\displaystyle \frac{t}{s}})}}$, which is a principal ideal. □

Note: It is easy to see that $B={\left((0,t)\right)}_i$ or $B={\left((-s,0)\right)}_i$ for any $s,t\in {\mathbb{R}}^{+}$, and $\left\{(0,0)\right\}={\left((0,0)\right)}_i$. Hence, by Theorem 4, we conclude that every ideal of B, being a closed set, is a principal ideal.

Theorem 5.

Let I be a prime ideal of a semiring B. Then, the following statements hold:

1. 

If I is a closed set, then $I=\left\{\right(0,0\left)\right\}$ or $I=\overline{I_{(-1,1)}}$.

2. 

If $I\setminus \left\{\right(0,0\left)\right\}$ is an open set, then $I=B^{*}\cup \left\{(0,0)\right\}$.

Proof. 

• Assume that I is any nontrivial proper ideal of B, being a closed set. Then, by Theorem 3, $I=\overline{I_{(-1,c)}}$, where $c\in \left(0,1\right]$.
  Case 1: $c=1$. Let $(-a,b),(-c,d)\in B$ such that $(-a,b)(-c,d)\in \overline{I_{(-1,1)}}$. Since $(-a,b)(-c,d)=(-ad-bc,ac+bd)$, we have $ad+bc=ac+bd$. If $d-c=0$, then $(-c,d)\in \overline{I_{(-1,1)}}$. If $d-c\ne 0$, then $a=b$. This implies that $(-a,b)\in \overline{I_{(-1,1)}}$. Hence, $\overline{I_{(-1,1)}}$ is a prime ideal.
  Case 2: $c\ne 1$. Note that $\left({\displaystyle -1,\frac{c}{1+ϵ}}\right)\notin \overline{I_{(-1,c)}}$ for any $ϵ>0$. So, $\left({\displaystyle -\frac{c}{1+ϵ},1}\right)\notin \overline{I_{(-1,c)}}$ since $\overline{I_{(-1,c)}}$ is an ideal. If we find that $ϵ>0$ such that

  $$\left(-1,{\displaystyle \frac{c}{1+ϵ}}\right)\left(-{\displaystyle \frac{c}{1+ϵ}},1\right)\in \overline{I_{(-1,c)}},$$

  then we can conclude that $\overline{I_{(-1,c)}}$ is not a prime ideal.
  Consider $\left({\displaystyle -1,\frac{c}{1+ϵ}}\right)\left({\displaystyle -\frac{c}{1+ϵ},1}\right)\in \overline{I_{(-1,c)}}$. This implies that

  $${\displaystyle \frac{c}{1}}\le {\displaystyle \frac{2c(1+ϵ)}{{(1+ϵ)}^2+c^2}}\le {\displaystyle \frac{1}{c}}.$$
  So, $c(1+2ϵ+{ϵ}^2+c^2)\le 2c(1+ϵ)$ and $2c^2(1+ϵ)\le 1+2ϵ+{ϵ}^2+c^2$. Then,

  $${ϵ}^2\le 1-c^2\mathrm{and}0\le {ϵ}^2+(2ϵ+1)(1-c^2).$$
  Thus, $0<ϵ\le \sqrt{1-c^2}$ because $0\le {ϵ}^2+(2ϵ+1)(1-c^2)$ is true for any $ϵ>0$ and $c\in (0,1]$. Hence, if we choose ${\displaystyle ϵ=\frac{1}{2}\sqrt{1-c^2}}$, then we have proven that $\overline{I_{(-1,c)}}$ is not a prime ideal.
  As a result, $\left\{\right(0,0\left)\right\}$ and $\overline{I_{(-1,1)}}$ are only two prime ideals of B, being closed sets.
• Since $I\setminus \left\{\right(0,0\left)\right\}$ is an open set, by Corollary 2, we conclude that $I=I_{(-1,c)}\cup \left\{(0,0)\right\}$ or $I=B^{*}\cup \left\{(0,0)\right\}$, where $0<c<1$. Likewise, the proof of 1. in Case 2 confirms that $I_{(-1,c)}\cup \left\{(0,0)\right\}$ is not a prime ideal. Applying the contrapositive of the definition of prime ideals, we conclude that $B^{*}\cup \left\{(0,0)\right\}$ is a prime ideal.

□

Theorem 6.

$B^{*}\cup \left\{(0,0)\right\}$ is a maximal ideal of a semiring B.

Proof. 

Let I be any ideal of a semiring B such that $B^{*}\cup \left\{(0,0)\right\}\subseteq I\subseteq B$.

• Case 1: There exists $(0,b)\in I$. To show that $I=B$, let $x>0$ be given. Then, $(0,x)=\left({\displaystyle 0,\frac{x}{b}}\right)(0,b)\in I$ and $(-x,0)=\left({\displaystyle -\frac{x}{b},0}\right)(0,b)\in I$ since I is an ideal. So, $I=B$.
• Case 2: There exists $(-a,0)\in I$. Then, $(0,a)=(-1,0)(-a,0)\in I$, and by Case 1, we obtain $I=B$.
• Case 3: For any $a,b>0$, $(-a,0)$ and $(0,b)$ are not elements in I. To show that $I=B^{*}\cup \left\{(0,0)\right\}$, let $(-x,y)\in I$. Since $x,y>0$, we have $(-x,y)\in B^{*}\subseteq B^{*}\cup \left\{(0,0)\right\}$. □

Throughout Theorems 7–9, we examine the smallest ideal ${\left(A\right)}_i$ of B containing any nonempty subset A as follows:

Theorem 7 presents a finite set A. Next, a nonempty compact set A is established in Theorem 8, and finally, the necessary and sufficient conditions to clarify the shape of ${\left(A\right)}_i$ for any nonempty closed subset A are stated in Theorem 9.

Theorem 7.

Let A be a subset of a semiring B. Then, the following statements hold:

1. 

For each $a,b\in {\mathbb{R}}^{+}$, if $A=\left\{\right(-a,b\left)\right\}$, then ${\left(A\right)}_i=\overline{I_{(-1,m)}}$, where $m=min\left\{{\displaystyle \frac{a}{b},\frac{b}{a}}\right\}$.

2. 

If A is a finite set, then ${\left(A\right)}_i=\overline{I_{(-1,m)}}$, where $m=min\left\{{\displaystyle \frac{s}{t},\frac{t}{s}|(-s,t)\in A}\right\}$.

Proof. 

• Assume $A=\left\{\right(-a,b\left)\right\}$, where $a,b\in {\mathbb{R}}^{+}$.
  Case 1: $a=b$. Since $(-a,a)\in A\subseteq {\left(A\right)}_i$ and ${\left(A\right)}_i$ is an ideal of B, we obtain

  $$(-x,x)=\left(0,{\displaystyle \frac{x}{a}}\right)(-a,a)\in {\left(A\right)}_i$$

  for each $(-x,x)\in \overline{I_{(-1,1)}}$. Thus, $\overline{I_{(-1,1)}}\subseteq {\left(A\right)}_i$. But, $(-a,a)\in \overline{I_{(-1,1)}}$, so $\overline{I_{(-1,1)}}$ is an ideal of B containing A. This implies that ${\left(A\right)}_i\subseteq \overline{I_{(-1,1)}}$.
  Case 2: $a\ne b$. We only show the case $a>b$. The other case can be proven similarly. Since $(-a,b)\in A\subseteq {\left(A\right)}_i$ and ${\left(A\right)}_i$ is an ideal of B, it follows that for any $(-x,y)\in L_1:=\left\{{\displaystyle (-x,y)\in B^{*}|y=-\frac{b}{a}(-x)}\right\}$, we have

  $$(-x,y)=\left(-x,{\displaystyle \frac{bx}{a}}\right)=\left(0,{\displaystyle \frac{x}{a}}\right)(-a,b)\in {\left(A\right)}_i.$$
  Thus, $L_1\subseteq {\left(A\right)}_i$. Next, let $L_2:=\left\{{\displaystyle (-x,y)\in B^{*}|y=-\frac{a}{b}(-x)}\right\}$. Then, we have $L_2=(-1,0)L_1\subseteq {\left(A\right)}_i$. Last, for any $(-x,y)\in \overline{I_{\left(-1,{\displaystyle \frac{b}{a}}\right)}}\setminus (L_1\cup L_2)$, we can write

  $$(-x,y)=\left(0,{\displaystyle \frac{x{a}^2-yab}{a^2-b^2}}\right)\left(-1,{\displaystyle \frac{b}{a}}\right)+\left(0,{\displaystyle \frac{yab-x{b}^2}{a^2-b^2}}\right)\left(-1,{\displaystyle \frac{a}{b}}\right)$$

  because $a>b$ and ${\displaystyle \frac{b}{a}x<y<\frac{a}{b}x}$. So, $\overline{I_{\left(-1,{\displaystyle \frac{b}{a}}\right)}}\subseteq {\left(A\right)}_i$. But, $(-a,b)\in \overline{I_{\left(-1,{\displaystyle \frac{b}{a}}\right)}}$, so $\overline{I_{\left(-1,{\displaystyle \frac{b}{a}}\right)}}$ is an ideal of B containing A. This implies that ${\left(A\right)}_i\subseteq \overline{I_{\left(-1,{\displaystyle \frac{b}{a}}\right)}}$.
• Assume that A is a finite subset of B. We only show the case $A=\left\{\right(-a,b),(-c,d\left)\right\}$. In other cases, there is a similar process. Let $m=min\left\{{\displaystyle \frac{s}{t},\frac{t}{s}|(-s,t)\in A}\right\}$. By 1. and Remark 1 (5.), we obtain $A\subseteq \overline{I_{(-1,m)}}$ and then ${\left(A\right)}_i\subseteq \overline{I_{(-1,m)}}$. Conversely, we separate two cases. If $min\left\{{\displaystyle \frac{a}{b},\frac{b}{a}}\right\}<min\left\{{\displaystyle \frac{c}{d},\frac{d}{c}}\right\}$, we obtain $\overline{I_{(-1,m)}}={\left(\left\{\right(-a,b\left)\right\}\right)}_i\subseteq {\left(A\right)}_i$. If $min\left\{{\displaystyle \frac{a}{b},\frac{b}{a}}\right\}>min\left\{{\displaystyle \frac{c}{d},\frac{d}{c}}\right\}$, similarly, $\overline{I_{(-1,m)}}\subseteq {\left(A\right)}_i.$

□

Note: It is evident that if $A=\left\{\right(0,0\left)\right\}$, then ${\left(A\right)}_i=\left\{(0,0)\right\}$, and if A has an element $(-s,0)$ or $(0,t)$ where $s,t>0$, then ${\left(A\right)}_i=B$, which is why we consider $A⊊B^{*}$ in the following theorems.

Theorem 8.

Let $A⊊B^{*}$ be a nonempty compact set. Define $A_S^{\prime }=A\cup (-1,0)A$, and $A_S$ is a part of $A_S^{\prime }$ under a line $y=-x$. Then, the following statements hold:

1. 

${\left(A_S\right)}_i=\overline{I_{(-s,t)}}$, where $t=inf\left\{y\right|(-x,y)\in A_S\}$ and $s=sup\left\{x\right|(-x,t)\in A_S\}$.

2. 

${\left(A_S\right)}_i={\left(A_S^{\prime }\right)}_i$.

3. 

${\left(A_S\right)}_i={\left(A\right)}_i$.

Proof. 

• Since A is a compact set, we have $(-s,t)\in A_S$. By Theorem 3, we have ${\left(A_S\right)}_i=\overline{I_{(-1,c)}}$, where $c\in \left(0,1\right]$. Actually, c is the infimum of the set $\left\{{\displaystyle \frac{y}{x}|(-x,y)\in {\left(A_S\right)}_i}\right\}$. Since the set $A_S$ is closed and bounded, we have $(-s,t)\in A_S\subseteq {\left(A_S\right)}_i$ and then ${\displaystyle c\le \frac{t}{s}}$. For any $(-a,b)\in \overline{I_{(-s,t)}}$, we can see that

  $$c\le {\displaystyle \frac{t}{s}}\le {\displaystyle \frac{b}{a}}\le {\displaystyle \frac{s}{t}}\le {\displaystyle \frac{1}{c}}.$$
  This implies that $(-a,b)\in \overline{I_{(-1,c)}}={\left(A_S\right)}_i$, so $\overline{I_{(-s,t)}}\subseteq {\left(A_S\right)}_i$. We now show that ${\left(A_S\right)}_i\subseteq \overline{I_{(-s,t)}}$. Let $(-a,b)\in A_S$. Without loss of generality, assume $a\ge b$. By the definition of t, we have $t\le b$. Let $L=\left\{(-x,y)\in B^{*}|y=\left({\displaystyle -\frac{b}{a}}\right)(-x)\right\}$. There exists $t_s>0$ such that $t_s=\left({\displaystyle -\frac{b}{a}}\right)(-s)$ and $t_s\ge t$. This implies that

  $${\displaystyle \frac{b}{a}}={\displaystyle \frac{t_s}{s}}\ge {\displaystyle \frac{t}{s}},$$

  and ${\displaystyle b\ge \frac{t}{s}a}$. Since $a\ge b$, we obtain ${\displaystyle \frac{b}{a}\le 1\le \frac{s}{t}}$, so ${\displaystyle b\ge \frac{s}{t}a}$. As a result, ${\displaystyle \frac{t}{s}a\le b\le \frac{s}{t}a}$ and then $(-a,b)\in \overline{I_{(-s,t)}}$. Thus, $\overline{I_{(-s,t)}}$ is an ideal containing $A_S$, and it follows that ${\left(A_S\right)}_i\subseteq \overline{I_{(-s,t)}}$.
• Since $A_S\subseteq A_S^{\prime }$, we obtain ${\left(A_S\right)}_i\subseteq {\left(A_S^{\prime }\right)}_i$. Conversely, let $(-a,b)\in A_S^{\prime }$.
  Case 1: $(-a,b)\in A$. If $(-a,b)$ is under the line $y=-x$, then $(-a,b)\in A_S\subseteq {\left(A_S\right)}_i$. If $(-a,b)$ is above the line $y=-x$, then $(-b,a)$ is under the line $y=-x$ and $(-b,a)\in A_S\subseteq {\left(A_S\right)}_i$. So, $(-a,b)=(-1,0)(-b,a)\in {\left(A_S\right)}_i$.
  Case 2: $(-a,b)\in (-1,0)A$. This implies that $(-b,a)\in A$. Similar to Case 1, we have $(-b,a)\in {\left(A_S\right)}_i$ and hence $(-a,b)=(-1,0)(-b,a)\in {\left(A_S\right)}_i$.
  As a result, $A_S^{\prime }\subseteq {\left(A_S\right)}_i$ and ${\left(A_S^{\prime }\right)}_i\subseteq {\left(A_S\right)}_i$.
• It suffices to show that ${\left(A\right)}_i={\left(A_S^{\prime }\right)}_i$. Since $A\subseteq A_S^{\prime }$, we obtain ${\left(A\right)}_i\subseteq {\left(A_S^{\prime }\right)}_i$. To show that $A_S^{\prime }\subseteq {\left(A\right)}_i$, let $(-a,b)\in A_S^{\prime }$. If $(-a,b)\in A$, then $(-a,b)\in {\left(A\right)}_i$. If $(-a,b)\in (-1,0)A$, then $(-b,a)\in A\subseteq {\left(A\right)}_i$ and $(-a,b)=(-1,0)(-b,a)\in {\left(A\right)}_i$. Thus, by the definition of ${\left(A_S^{\prime }\right)}_i$, we conclude that ${\left(A_S^{\prime }\right)}_i\subseteq {\left(A\right)}_i$.

□

Theorem 9.

Let $A\ne \left\{\right(0,0\left)\right\}$ be a nonempty closed subset of $B^{*}$. Define $A_S^{\prime }=A\cup (-1,0)A$, and $A_S$ is a part of $A_S^{\prime }$ under the line $y=-x$. For each $m\in (0,1]$, let

$$T_m=\{(-x,y)\in B^{*}|y\le (-m)(-x)\}.$$

Then, the following statements hold:

1. 

$A_s\cap T_m\ne \varnothing$ for any m if and only if ${\left(A\right)}_i=B^{*}\cup \left\{(0,0)\right\}$.

2. 

$A_s\cap T_m=\varnothing$ for some m if and only if ${\left(A\right)}_i=\overline{I_{(-s,t)}}$ for some positive real numbers s and t.

3. 

$A\subseteq T_1$ if and only if ${\left(A\right)}_i=\overline{I_{(-1,1)}}$.

Proof. 

• Suppose $A_s\cap T_m\ne \varnothing$ for all m. It suffices to show that $B^{*}\subseteq {\left(A\right)}_i$. Let $(-a,b)\in B^{*}$. If $a\ge b$, then there exists $(-a_1,b_1)\in A_s\cap T_{{\displaystyle \frac{b}{a}}}$. Since $A_s\subseteq {\left(A\right)}_i$ and $(-a_1,b_1)\in {\left(A\right)}_i$ is an ideal, by Theorem 7 (1.), we obtain $\overline{I_{(-a_1,b_1)}}\subseteq {\left(A\right)}_i$. Since $(-a_1,b_1)\in T_{{\displaystyle \frac{b}{a}}}$, we obtain ${\displaystyle \frac{b_1}{a_1}\le \frac{b}{a}\le \frac{a}{b}\le \frac{a_1}{b_1}}$ and then $(-a,b)\in \overline{I_{(-a,b)}}\subseteq \overline{I_{(-a_1,b_1)}}\subseteq {\left(A\right)}_i$. If $a<b$, then $(-b,a)\in {\left(A\right)}_i$. By the fact that $(-a,b)=(-1,0)(-b,a)$, we also have $(-a,b)\in {\left(A\right)}_i$. Conversely, we prove by contrapositive. Suppose $A_s\cap T_m=\varnothing$ for some $m\in (0,1]$. Let $(-a,b)\in A$. If $a\ge b$, then $(-a,b)\in A_s$ and then $(-a,b)\notin T_m$. Since ${\displaystyle m<1<\frac{a}{b}<\frac{1}{m}}$, we have $(-b,a)\in \overline{I_{(-1,m)}}$ and then $(-a,b)\in \overline{I_{(-1,m)}}$. If $a<b$, then $(-b,a)\in A_s$ and $(-b,a)\notin T_m$. In the same manner, we have $(-a,b)\in \overline{I_{(-1,m)}}$. This implies that $A\subseteq \overline{I_{(-1,m)}}$. Since $\overline{I_{(-1,m)}}$ is an ideal containing A and $\overline{I_{(-1,m)}}$ is a proper subset of $B^{*}\cup \left\{(0,0)\right\}$, we conclude that ${\left(A\right)}_i\ne B^{*}\cup \left\{(0,0)\right\}$.
• Suppose there exists $(-s,t)\in B^{*}$ such that ${\left(A\right)}_i=\overline{I_{(-s,t)}}$. Since $\overline{I_{(-s,t)}}⊊B^{*}\cup \left\{(0,0)\right\}$, we have $A_s\cap T_m=\varnothing$ for some $m>0$ by 1. Conversely, suppose that there exists $m_0>0$ such that $A_s\cap T_{m_0}=\varnothing$. By 1, we obtain ${\left(A\right)}_i\ne B^{*}\cup \left\{(0,0)\right\}$. Let

  $$M=\{m\in (0,1]|A_s\cap T_m=\varnothing \}$$

  and $m=supM$. Then, we show that ${\left(A\right)}_i=\overline{I_{(-1,m)}}$. Let $(-x,y)\in \overline{I_{(-1,m)}}$. Given the property of the ideal $\overline{I_{(-1,m)}}$, we assume that $x\ge y$. It follows that ${\displaystyle m<\frac{y}{x}\le 1}$. If $A_s\cap T_{{\displaystyle \frac{y}{x}}}=\varnothing$, then ${\displaystyle \frac{y}{x}}\in M$, which is a contradiction. Thus, $A_s\cap T_{{\displaystyle \frac{y}{x}}}\ne \varnothing$, implying there exists $(-a,b)$ such that $a\ge b$. This implies that $(-a,b)\in A_s\subseteq {\left(A\right)}_i$. Since $(-a,b)\in T_{{\displaystyle \frac{y}{x}}}$, we have ${\displaystyle \frac{b}{a}\le \frac{y}{x}\le 1}$. Thus, $(-x,y)\in \overline{I_{(-x,y)}}\subseteq \overline{I_{(-a,b)}}\subseteq {\left(A\right)}_i$. To show that ${\left(A\right)}_i\subseteq \overline{I_{(-1,m)}}$, by Theorem 8 (2. and 3.), it is enough to show that $A_s\subseteq \overline{I_{(-1,m)}}$. Let $(-x,y)\in A_s$ and $x\ge y$. If $(-x,y)\notin \overline{I_{(-1,m)}}$, then ${\displaystyle \frac{y}{x}<m\le 1}$ and then $y<(-m)(-x)$. This implies that $(-x,y)\in T_m$, so $A_s\cap T_m\ne \varnothing$, which is a contradiction.
• We only prove the sufficient part. Clearly, ${\left(A\right)}_i\subseteq \overline{I_{(-1,1)}}$ by assumption. Let $(-x,x)\in \overline{I_{(-1,1)}}$. Since $A\ne \varnothing$, let $(-a,a)\in A\subseteq {\left(A\right)}_i$. Then, $(-x,x)=\left({\displaystyle 0,\frac{x}{a}}\right)(-a,a)\in {\left(A\right)}_i$.

□

3.2. Proof Utilizing the Isomorphism Approach

First, we define a set

$$B^{\prime }:=\{(x,y)\in {\mathbb{R}}^2|x\ge |y\left|\right\}$$

and a map $T:B\to B^{\prime }$ (see Figure 7), where for each $(-a,b)\in B$,

$$T(-a,b)=(b+a,b-a).$$

These additions were recommended by a reviewer to support our findings.

Then, $B^{\prime }$ is a commutative semiring with identity $(1,1)$ and zero $(0,0)$ under pointwise addition and multiplication, and T is an isomorphism of semirings, as shown in Proposition 2.

In this section, the reviewer’s suggestion that the set $B^{\prime }$ and the mapping T results in a reduction in the complexity of multiplication on B. To enhance clarity, we have to eliminate certain details from the proofs of our main results, Theorems 3–6. For brevity, refer to Theorems 10–13.

Proposition 2. 

Let B and $B^{\prime }$ be the above-defined semirings. If a map T is defined as above, then T is an isomorphism of semirings.

Proof. 

Clearly, $T(0,0)=(0,0)$ and $T(0,1)=(1,1)$. A map T is a homomorphism because for any $(-a,b),(-c,d)\in B$,

$$\begin{array}{cc}\hfill T(-a,b)+T(-c,d)& =(b+a,b-a)+(d+c,d-c)\hfill \\ \hfill & =(b+a+d+c,b+d-a-c)\hfill \\ \hfill & =T(-a-c,b+d)=T\left(\right(-a,b)+(-c,d\left)\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill T(-a,b)T(-c,d)& =(b+a,b-a)(d+c,d-c)\hfill \\ \hfill & =\left(\right(b+a\left)\right(d+c),(b-a\left)\right(d-c\left)\right)\hfill \\ \hfill & =(bd+bc+ad+ac,bd-bc-ad+ac)\hfill \\ \hfill & =T(-bc-ad,ac+bd)=T\left(\right(-a,b\left)\right(-c,d\left)\right).\hfill \end{array}$$

To show that T is 1-1, let $(-a,b),(-c,d)\in B$ such that $T(-a,b)=T(-c,d)$. Then, $b+a=d+c$ and $b-a=d-c$. This implies that $a=c$ and $b=d$; hence, $(-a,b)=(-c,d)$. Lastly, to show that T is on $B^{\prime }$, let $(x,y)\in B^{\prime }$. Then, $x\ge \left|y\right|$. Set ${\displaystyle a=\frac{x-y}{2}}$ and ${\displaystyle b=\frac{x+y}{2}}$. Then, $(-a,b)\in B$ and $T(-a,b)=(b+a,b-a)=(x,y)$. As a result, T is an isomorphism of semirings. □

Next, we introduce the notations that play a significant role in our results for this section. For $r\in [0,1]$, we let

$$I_r=\left\{\begin{array}{cc}\{(x,y)\in {\mathbb{R}}^2|rx\ge \left|y\right|\}\hfill & \mathrm{if}r\ne 0\hfill \\ \{(x,0)\in {\mathbb{R}}^2|x\ge 0\}\hfill & \mathrm{if}r=0,\hfill \end{array}\right.$$

and

$$I_r^{*}=\left\{\begin{array}{cc}\{(x,y)\in {\mathbb{R}}^2|rx>\left|y\right|\}\hfill & \mathrm{if}r\ne 0\hfill \\ \{(x,0)\in {\mathbb{R}}^2|x>0\}\hfill & \mathrm{if}r=0,\hfill \end{array}\right.$$

Remark 2.

From Proposition 2, we know that T is an isomorphism of semirings. The following sets are isomorphic:

• $\{(-a,0)\in {\mathbb{R}}^2|a\ge 0\}\cong \{(a,b)\in B^{\prime }|a=-b\}$;
• $\{(0,b)\in {\mathbb{R}}^2|b\ge 0\}\cong \{(a,b)\in B^{\prime }|a=b\}$;
• $\overline{I_{(-1,1)}}\cong I_0$;
• $\overline{I_{(-s,t)}}\cong I_r$; $I_{(-s,t)}\cong I_r^{*}$, where ${\displaystyle r=\frac{s-t}{s+t}}$.

Theorem 10.

Let I be any nonempty subset of $B^{\prime }$.

1. 

For a closed set I of $B^{\prime }$, I is an ideal on $B^{\prime }$ if and only if there exists $r\in (0,1]$ such that $I=I_r$ or $I=I_0$.

2. 

For an open set $I\setminus \left\{\right(0,0\left)\right\}$ of $B^{\prime }$, I is an ideal on $B^{\prime }$ if and only if there exists $r\in (0,1]$ such that $I=I_r^{*}\cup \left\{(0,0)\right\}$.

Proof. 

• Clearly, $I_0$ is a closed set and an ideal of $B^{\prime }$, and $I_r$ is a closed set of $B^{\prime }$. To show that $I_r$ is an ideal of $B^{\prime }$, let $(x,y),(a,b)\in I_r$. Then, $rx\ge \left|y\right|$ and $ra\ge \left|b\right|$. So, $r(x+a)\ge \left|y\right|+\left|b\right|\ge |y+b|$. This implies that $(x,y)+(a,b)=(x+a,y+b)\in I_r$. Let $(u,v)\in B^{\prime }$. Since $rx\ge \left|y\right|$ and $u\ge \left|v\right|$, we have $r\left(xu\right)\ge \left|yv\right|$ and then $(x,y)(u,v)=(xu,yv)\in I_r$. Now, we shall proceed to prove the sufficient condition. Assume that I is an ideal on $B^{\prime }$. If $I\subseteq I_0$, then it is easy to see that $I=I_0$. Now, we can assume that there exists $(a,b)\in I$ such that $a>0$ and $b\ne 0$ since I is an ideal. Let $R=\left\{{\displaystyle \frac{\left|y\right|}{x}|(x,y)\in I}\right\}$. Then, $R\ne \varnothing$. Since $x\ge \left|y\right|\ge 0$ for any $(x,y)\in I\subseteq B^{\prime }$, we conclude that 1 is an upper bound of the set R. So, we can let $r=supR$. Since $(a,b)\in I$ and ${\displaystyle \frac{\left|b\right|}{a}>0}$, it follows that r has to be an element in $(0,1]$. Since I is a closed set, there exists $(c,d)\in I$ such that ${\displaystyle \frac{\left|d\right|}{c}=r}$. We now show that $I=I_r$. Since $r=supR$, ${\displaystyle \frac{\left|y\right|}{x}\le r}$ for any $(x,y)\in I$. That is, $I\subseteq I_r$. Conversely, for any $(x,y)\in {\mathbb{R}}^2$ such that $rx\ge \left|y\right|$, we obtain ${\displaystyle \left|y\right|\le rx=\frac{\left|d\right|}{c}x}$ and then ${\displaystyle \frac{\left|y\right|}{\left|d\right|}\le \frac{x}{c}}$. This implies that ${\displaystyle \left(\frac{x}{c},\frac{y}{d}\right)\in B^{\prime }}$. Thus, ${\displaystyle (x,y)=\left(\frac{x}{c},\frac{y}{d}\right)(c,d)\in I}$.
• The outline of the proof is analogous to that of the closed set version.

□

Theorem 11.

For each $(s,t)\in B^{\prime }$,

$${\left((s,t)\right)}_i=\left\{\begin{array}{cc}\left\{\right(0,0\left)\right\}\hfill & \mathrm{if}s=0\hfill \\ I_{{\displaystyle \frac{\left|t\right|}{s}}}\hfill & \mathrm{if}s>0.\hfill \end{array}\right.$$

Proof. 

Let $(s,t)\in B^{\prime }$. If $s=0$, then t must be equal to zero and ${\left((s,t)\right)}_i=\left\{(0,0)\right\}$. If $s>0$ and $t=0$, it follows that ${\left((s,t)\right)}_i=I_0$. Suppose $s>0$ and $t\ne 0$. We show that ${\left((s,t)\right)}_i=I_{{\displaystyle \frac{\left|t\right|}{s}}}$. Since ${\displaystyle \frac{\left|t\right|}{s}s\ge \left|t\right|}$, we obtain $(s,t)\in I_{{\displaystyle \frac{\left|t\right|}{s}}}$. So, ${\left((s,t)\right)}_i\subseteq I_{{\displaystyle \frac{\left|t\right|}{s}}}$. For each $(x,y)\in I_{{\displaystyle \frac{\left|t\right|}{s}}}$, we have ${\displaystyle \frac{\left|t\right|}{s}x\ge \left|y\right|}$, that is, $\left({\displaystyle \frac{x}{s},\frac{y}{t}}\right)\in B^{\prime }$. Thus, $(x,y)=\left({\displaystyle \frac{x}{s},\frac{y}{t}}\right)(s,t)\in {\left((s,t)\right)}_i$, so $I_{{\displaystyle \frac{\left|t\right|}{s}}}\subseteq {\left((s,t)\right)}_i$. □

Remark 3.

In actuality, the constant $r\in (0,1]$ in Theorem 10 can be represented as ${\displaystyle r=\frac{\left|d\right|}{c}}$, where $(c,d)$ is the boundary point of I if I is a closed set. According to Theorem 10, for any closed ideal I of $B^{\prime }$, we can conclude that $I=I_{{\displaystyle \frac{\left|d\right|}{c}}}$ for some $(c,d)\in I$, and then $I={\left((c,d)\right)}_i$ by Theorem 11. This means that I is a principal ideal. Hence, every ideal of $B^{\prime }$, being a closed set, is a principal ideal.

Theorem 12.

Let I be a prime ideal of a semiring $B^{\prime }$. Then, the following statements hold:

1. 

If I is a closed set, then $I=\left\{\right(0,0\left)\right\}$ or $I=I_0$.

2. 

If $I\setminus \left\{\right(0,0\left)\right\}$ is an open set, then $I=I_1^{*}\cup \left\{(0,0)\right\}$.

Proof. 

To show that $I_r$ and $I_r^{*}$ are not prime ideals for any $r\in (0,1)$, let $r\in (0,1)$. Then, $(r-1)(r-4)>0$, and it follows that ${\displaystyle 0<\frac{\sqrt{r}-r}{2}<1-r}$. Let ${\displaystyle ϵ=\frac{\sqrt{r}-r}{2}}$. Then, $(1,r+ϵ)\notin I_r^{*}$ and $(1,r+ϵ)\notin I_r$ since $r+ϵ>r$. But,

$$(1,r+ϵ)(1,r+ϵ)=(1,{(r+ϵ)}^2)\in I_R^{*}\subseteq I_r$$

because ${\displaystyle {(r+ϵ)}^2=\frac{{(r+\sqrt{r})}^2}{4}=\frac{r^2+2r\sqrt{r}+r}{4}<\frac{r+2r\left(1\right)+r}{4}=r}$. □

Theorem 13.

$I_1^{*}\cup \left\{(0,0)\right\}$ is a maximal ideal of a semiring $B^{\prime }$.

Proof. 

Let I be an ideal such that $I_1^{*}\cup \left\{(0,0)\right\}\subseteq I\subseteq B^{\prime }$. Suppose $I⊈I_1^{*}\cup \left\{(0,0)\right\}$. Then, there exists $x>0$ such that $(x,x)\in I\setminus (I_1^{*}\cup \left\{(0,0)\right\})$ or $(x,-x)\in I\setminus (I_1^{*}\cup \left\{(0,0)\right\})$. Since $B^{\prime }=I_1={\left((x,x)\right)}_i={\left((x,-x)\right)}_i\subseteq I$, we have $B^{\prime }=I$. Hence, $I_1^{*}\cup \left\{(0,0)\right\}$ is a maximal ideal of $B^{\prime }$. □

The following table (see

Table 3. Comparison between main results in Section 3.1 and Section 3.2.

Type	In ${\mathit{B}}^{\prime }$	In B
Principal Ideal	$\left\{\right(0,0\left)\right\}$	$\left\{\right(0,0\left)\right\}$
	$I_{{\displaystyle \frac{\left|t\right|}{s}}}$ where $s\ge \left|t\right|$	$\overline{I_{\left(-{\displaystyle \frac{s+\left|t\right|}{2}},{\displaystyle \frac{s-\left|t\right|}{2}}\right)}}$ where $s\ge \left|t\right|$
Prime Ideal	$\left\{\right(0,0\left)\right\}$	$\left\{\right(0,0\left)\right\}$
	$I_0$	$\overline{I_{(-1,1)}}$
	$I_1^{*}\cup \left\{(0,0)\right\}$	$B^{*}\cup \left\{(0,0)\right\}$
Maximal Ideal	$I_1^{*}\cup \left\{(0,0)\right\}$	$B^{*}\cup \left\{(0,0)\right\}$

) presents a comparison of the results obtained from Theorems 4–6 in Section 3.1 and Theorems 11–13.

4. Discussion and Conclusions

Briefly, on a semiring $(B,+,·)$, where $B=\left\{(-x,y)\in {\mathbb{R}}^2|x,y\ge 0\right\}$, there are only six possible forms for the ideal I (see Figure 4 and Figure 5), as follows:

• A singleton set $\left\{\right(0,0\left)\right\}$;
• A straight line $y=-x$;
• A V-shaped region $\overline{I_{(-s,t)}}=\left\{(-x,y)\in {\mathbb{R}}^2|{\displaystyle \frac{t}{s}}x\le y\le {\displaystyle \frac{s}{t}}x\right\}$, where $s,t\in {\mathbb{R}}^{+}$ such that $s>t$;
• A V-shaped region $I_{(-s,t)}\cup \left\{(0,0)\right\}=\left\{(-x,y)\in B|{\displaystyle \frac{t}{s}}x<y<{\displaystyle \frac{s}{t}}x\right\}\cup \left\{(0,0)\right\}$, where $s,t\in {\mathbb{R}}^{+}$ such that $s>t$;
• $B^{*}\cup \left\{(0,0)\right\}$, where $B^{*}=\left\{(-x,y)\in {\mathbb{R}}^2|x,y>0\right\}$;
• A whole space B.

All of them are symmetric with the graph $y=-x$. In particular, we also guarantee that every ideal of B, being a closed set, is a principal ideal. The sets $\left\{\right(0,0\left)\right\}$ and $\overline{I_{(-1,1)}}$ are only two prime ideals, being closed sets. And, $B^{*}\cup \left\{(0,0)\right\}$ is the only maximal ideal of a semiring B.

For more details, we study how the set

$$\overline{I_{(-s,t)}}=\left\{(-x,y)\in B|{\displaystyle \frac{t}{s}}x\le y\le {\displaystyle \frac{s}{t}}x\right\},$$

where $(-s,t)\in B$ and $s\ge t$, relates to the ideal in a semiring B. Then, we find the following:

• All nontrivial proper ideals of B, being a closed set, are in the form $\overline{I_{(-1,c)}}$ for some $c\in (0,1]$.
• Every nontrivial proper ideal of B, being a closed set, is a principal ideal.
• All possible prime ideals are classified, that is, one of the prime ideals of B is $\overline{I_{(-1,1)}}$.
• For any nonempty proper compact set A of $B^{*}$, there exists a point $(-s,t)\in B$ such that ${\left(A\right)}_i=\overline{I_{(-s,t)}}$, where ${\left(A\right)}_i$ is the ideal of B generated by A.
• The necessary and sufficient conditions for ${\left(A\right)}_i=B^{*}\cup \left\{(0,0)\right\}$, ${\left(A\right)}_i=\overline{I_{(-s,t)}}$ for some $(-s,t)\in B^{*}$, and ${\left(A\right)}_i=\overline{I_{(-1,1)}}$ are distributed, where $A\ne \left\{\right(0,0\left)\right\}$ is any nonempty closed subset of $B^{*}$.

Actually, we envision a V-shape of ideal I in the bipolar set B from the multiplicative definition. We consider such an element of B as a vector in ${\mathbb{R}}^2$ space. In order to obtain a clear conceptual grasp, if we let $(-3,4)\in I$, then each vector lies on the straight line ${\displaystyle y=-\frac{4}{3}x}$ contained in I, say $L_1$, because $(-3b,4b)=(0,b)(-3,4)\in I$ for any $b\ge 0$. In addition, $(-4,3)=(-1,0)(-3,4)\in I$. Similarly, we observe that each vector lies on the straight line ${\displaystyle y=-\frac{3}{4}x}$ contained in I, say $L_2$. Thus, two straight lines form the boundary of a V-shape of ideal I containing a vector $(-3,4)$. Moreover, for any $(-a,b)\in B$, the result of $(-a,b)(-3,4)=(0,a)(-4,3)+(0,b)(-3,4)$ is a sum of two vectors, where one is in the straight line $L_1$ and another lies on the straight line $L_2$. As a result, $(-a,b)(-3,4)$ must be a vector between $L_1$ and $L_2$. This means that it is contained in a region with two straight lines $L_1$ and $L_2$ that are the boundaries of a V-shape.

In other words, we start with one point and we end with one V-shaped region corresponding to that point. This convinces us that the ideal must be the arbitrary union of V-shaped regions that originate from any point in I. Given the character of the V-shape, it forms a chain with a partially ordered relation “⊆”, ensuring it has a maximal element by Zorn’s lemma. Lastly, we provide more details: the maximal element is $\overline{I_{(-1,c)}}$, where $c=inf\left\{{\displaystyle \frac{y}{x}}|(-x,y)\in I\right\}$. In addition, we examine the structure of I on the semigroup $(B,+)$. The result in every ideal must be an arbitrary union of sets of rectangular shapes indexed by some set. So, we would like to move on from the concept of ideals to the theory of modules.

Moreover, in Section 3.2, we transform the set B by applying the semiring isomorphism $T(-a,b)=(b+a,b-a)$ so that $B\cong B^{\prime }$, where $B^{\prime }=\{(x,y)\in {\mathbb{R}}^2|x\ge |y\left|\right\}$. The main differences between $(B,+,·)$ and $(B^{\prime },+,·)$ lie in the multiplicative operations of B and $B^{\prime }$. So, we can avoid the complicated effect of multiplication on B—$(-a,b)(-c,d)=(-ad-bc,ac+bd)$ for $(-a,b),(-c,d)\in B$—compared to the easier multiplication on $B^{\prime }$—$(a,b)(c,d)=(ac,bd)$ for $(a,b),(c,d)\in B^{\prime }$. This streamlines many details in the proofs of our main results, Theorems 3–6, making them neater (see Theorems 10–13).