On k-Noetherian and k-Artinian Semirings

TL;DR

This paper explores the properties of left k-Noetherian and k-Artinian semirings, providing characterizations via i-injective semimodules and extending classical theorems to the semiring context.

Contribution

It introduces new characterizations of k-Noetherian and k-Artinian semirings using i-injective semimodules and presents a partial version of the Bass-Papp Theorem for semirings.

Findings

Characterization of k-Noetherian and k-Artinian semirings

Extension of Bass-Papp Theorem to semirings

Examples and counterexamples illustrating main results

Abstract

We investigate left k-Noetherian and left k-Artinian semirings. We characterize such semirings using i-injective semimodules. We prove in particular, a partial version of the celebrated Bass-Papp Theorem for semiring. We illustrate our main results by examples and counter examples.

Equations

$$a\cdot 0$$

$$a(b+c)$$

$$K^{+}(M)=\{x\in M\mid x+y=x+z⟹y=z\text{ for any }y,z\in M\}.$$

$$x\oplus y=-ln(e^{-x}+e^{-y})$$

$$B(n,i):=(B(n,i),\oplus ,0,\odot ,1),$$

$$\overline{L}:=\{m\in M\mid m+\ell ={\ell }^{\prime }\text{ for some }\ell ,{\ell }^{\prime }\in L\}.$$

$$\mathrm{Comp}(T):=\{t\in T\mid \exists \tilde{t}\in T\text{ with }t+\tilde{t}=1_T\text{ and }t\tilde{t}=0_T=\tilde{t}t\}.$$

$$L\stackrel{f}{⟶}M\stackrel{g}{⟶}N$$

$$\cdots \to M_{i-1}\stackrel{f_{i-1}}{\to }{M}_i\stackrel{f_i}{\to }{M}_{i+1}\stackrel{f_{i+1}}{\to }{M}_{i+2}\to \cdots$$

$$0⟶L\stackrel{f}{⟶}M\stackrel{g}{⟶}N⟶0.$$

$$N=L\oplus (K\cap N).$$

$$E_{1}=\left\{\left[{\begin{array}[]{cc}a&0\\ b&0\end{array}}\right]|\text{ }a,b\in\mathbb{R}^{+}\right\}\text{ and }E_{2}=\left\{\left[{\begin{array}[]{cc}0&c\\ 0&d\end{array}}\right]|\text{ }c,d\in\mathbb{R}^{+}\right\}$$

$$N_{\geq 1}:=\left\{\left[{\begin{array}[]{cc}a&c\\ b&d\end{array}}\right]\text{ }|\text{ }a\leq c,b\leq d,a,b,c,d\in\mathbb{R}^{+}\right\}.$$

$$N_{\ge 1}\cap (E_1\oplus E_2)\ne (N_{\ge 1}\cap E1)\oplus (N_{\ge 1}\cap E_2).$$

$$\mathcal{I}:=\{L{}_{S}N|L\text{ is a subtractive subsemimodule of }M\}.$$

$$L_0⫋L_1⫋L_2⫋\cdots ⫋L_k⫋L_{k+1}⫋\cdots$$

$$I_1⫌I_2⫌\cdots ⫌I_k⫌I_{k+1}⫌\cdots ,$$

$$E_1$$

$$E_2$$

$$N_r$$

$$\left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]\left[{\begin{array}[]{cc}a&0\\ b&0\end{array}}\right]+\left[{\begin{array}[]{cc}c&0\\ d&0\end{array}}\right]=\left[{\begin{array}[]{cc}pa+qb+c&0\\ ra+sb+d&0\end{array}}\right]\in E_{1}.$$

$$\left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]+\left[{\begin{array}[]{cc}a&0\\ b&0\end{array}}\right]=\left[{\begin{array}[]{cc}c&0\\ d&0\end{array}}\right]$$

$$\left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]\left[{\begin{array}[]{cc}ra&a\\ rb&b\end{array}}\right]+\left[{\begin{array}[]{cc}rc&c\\ rd&d\end{array}}\right]=\left[{\begin{array}[]{cc}r(ka+\ell b+c)&ka+\ell b+c\\ r(ma+nb+d)&ma+nb+d\end{array}}\right]\in N_{r}.$$

$$\left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]+\left[{\begin{array}[]{cc}ra&a\\ rb&b\end{array}}\right]=\left[{\begin{array}[]{cc}rc&c\\ rd&d\end{array}}\right],$$

$$\left[{\begin{array}[]{cc}p&0\\ r&0\end{array}}\right]+\left[{\begin{array}[]{cc}0&q\\ 0&s\end{array}}\right]=\left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]\in I$$

$$\left[{\begin{array}[]{cc}0&0\\ 0&1\end{array}}\right]=\left[{\begin{array}[]{cc}0&0\\ 1/q&0\end{array}}\right]\left[{\begin{array}[]{cc}0&q\\ 0&s\end{array}}\right].$$

$$\left[{\begin{array}[]{cc}0&0\\ 0&1\end{array}}\right]=\left[{\begin{array}[]{cc}0&0\\ 0&1/s\end{array}}\right]\left[{\begin{array}[]{cc}0&q\\ 0&s\end{array}}\right].$$

$$\left[{\begin{array}[]{cc}p&0\\ q&0\end{array}}\right]+\left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]=\left[{\begin{array}[]{cc}r\ell&\ell\\ rn&n\end{array}}\right]\in I$$

$$\left[{\begin{array}[]{cc}p&0\\ 0&q\end{array}}\right]+\left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]=\left[{\begin{array}[]{cc}r\ell&\ell\\ m&m/r\end{array}}\right]\in I.$$

$$\left[{\begin{array}[]{cc}1&0\\ 0&0\end{array}}\right]=\left[{\begin{array}[]{cc}1/p&0\\ 0&0\end{array}}\right]\left[{\begin{array}[]{cc}p&0\\ q&0\end{array}}\right]$$

$$\left[{\begin{array}[]{cc}1&0\\ 0&0\end{array}}\right]=\left[{\begin{array}[]{cc}1/p&0\\ 0&0\end{array}}\right]\left[{\begin{array}[]{cc}p&0\\ 0&q\end{array}}\right].$$

$$\left[{\begin{array}[]{cc}1/k&0\\ 0&0\end{array}}\right]\left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]=\left[{\begin{array}[]{cc}1&\ell/k\\ 0&0\end{array}}\right]$$

Full text

Introduction

Semirings are, roughly, rings not necessarily with subtraction, and generalize both rings and distributive bounded lattices. Semirings, and their semimodules (defined, roughly, as modules not necessarily with subtraction), have many important applications in several aspects of Computer Science and Mathematics, e.g., Automata Theory [HW1998], Tropical Geometry [Gla2002] and Idempotent Analysis [LM2005]. Our main reference for semirings and their semimodules is Golan’s book [Gol1999] and for rings and modules Wisbauer’s book [Wis1991].

Left (right) Noetherian rings, whose lattices of left (right) ideals satisfy the Ascending Chain Condition, are well studied due to the role it plays in simplifying the ideal structure such rings. On the other hand, left (right) Artinian rings, whose lattices of left (right) ideals satisfy the Descending Chain Condition, generalize simultaneously finite rings and rings that are finite-dimensional vector spaces over fields. Several properties of left (right) modules are valid only over rings with the ACC or the DCC. Some of these properties characterize such rings, e.g. the closure of the class of left (right) injective modules under arbitrary direct sums characterizes left (right) Noetherian rings [Rot2009, 3.39], and a ring $R$ is left (right) Artinian if and only if every finitely generated left (right) $R$-module is finitely cogenerated [Wis1991, 31.4].

A left (right) ideal $I$ of a semiring $S$ is called a $k$-ideal, iff $I\leq_{S}S$ is subtractive [Hen1958] (equivalently, $I=Ker(S\overset{\pi_{I}}{\longrightarrow}S/I),$ where $\pi_{I}$ is the canonical projection). In this paper, we consider the so called left $k$-Noetherian semirings (left $k$-Artinian semirings), whose lattice of subtractive left ideals satisfies the ACC (DCC). We generalize several results known for left Noetherian (left Artinian) rings to left $k$-Noetherian (left $k$-Artinian) semirings.

The paper is divided into two sections.

In Section 1, we collect the basic definitions, examples and preliminaries used in this paper. In particular, we recall the definitions and basic properties of *exact sequences *introduced by the first author Abuhlail [Abu2014].

In Section 2, we investigate left $k$-Noetherian (resp., left $k$-Artinian) semirings, i.e. semirings satisfying the ACC (resp., the DCC) on left $k$-ideals. In Example 2.10, we show that $S:=M_{2}(\mathbb{R}^{+})$ is left $k$-Noetherian but not left Noetherian, and is left $k$-Artinian but not left Artinian. In Theorem 2.13, we show that if every subtractive left ideal of a semiring $S$ is a direct summand, then $S$ is left $k$-Artinian and left $k$-Noetherian. In Theorem 2.19, we provide a partial version of the celebrated Bass-Papp Theorem for semirings: we show that if $S$ is a semiring with enough left $S$-$i$-injective semimodules and every direct sum of $S$-$i$-injective left $S$-semimodules is $S$-$i$-injective, then $S$ is left $k$-Noetherian.

1 Preliminaries

In this section, we provide the basic definitions and preliminaries used in this work. Any notions from the theory of semirings and semimodules that are not defined here can be found in our main reference [Gol1999]. We refer to [Wis1991] for the foundations of the theory of module and rings.

Definition 1.1**.**

([Gol1999]) A semiring is a datum $(S,+,0,\cdot,1)$ consisting of a commutative monoid $(S,+,0)$ and a monoid $(S,\cdot,1)$ such that $0\neq 1$ and

[TABLE]

1.2**.**

[Gol1999] Let $S$ and $T$ be semirings. The categories ${}_{S}\mathbf{SM}$ of left $S$-semimodules with arrows the $S$-linear maps, $\mathbf{SM}_{T}$ of right $S$-semimodules with arrows the $T$-linear maps, and ${}_{S}\mathbf{SM}_{T}$ of $(S,T)$-bisemimodules are defined in the usual way (as for modules and bimodules over rings). For a left $S$-semimodule $M,$ we write $L\leq_{S}M$ to indicate that $L$ is an $S$-subsemimodule of $M.$

Definitions 1.3**.**

([Gol1999]) Let $(S,+,0,\cdot,1)$ be a semiring.

• •

If the monoid $(S,\cdot,1)$ is commutative, we say that $S$ is a commutative semiring.

• •

We say that the semiring $S$ is additively idempotent, iff $s+s=s$ for every $s\in S.$

• •

The set of *cancellative elements of *a left $S$-semimodules $M$ is defined as

[TABLE]

We say that $M$ is a cancellative semimodule, iff $K^{+}(M)=M.$

Examples 1.4*.*

([Gol1999])

• •

Every ring is a cancellative semiring.

• •

Any distributive bounded lattice $\mathcal{L}=(L,\vee,1,\wedge,0)$ is a additively idempotent commutative semiring.

• •

The set $(\mathbb{Z}^{+},+,0,\cdot,1)$ (resp. $(\mathbb{Q}^{+},+,0,\cdot,1),$ $(\mathbb{Q}^{+},+,0,\cdot,1)$) of non-negative integers (resp. non-negative rational numbers, non-negative real numbers) is a cancellative commutative semiring which is not a ring.

• •

$M_{n}(S),$ the set of all $n\times n$ matrices over a semiring $S,$ is a semiring.

• •

$\mathbb{B}:=\{0,1\}$ with $1+1=1,$ is a an additively idempotent commutative semiring called the Boolean semiring.

• •

The max-plus algebra $\mathbb{R}_{\max,+}:=(\mathbb{R}\cup\{-\infty\},\max,-\infty,+,0)$ is a semiring.

• •

The log algebra $(\mathbb{R}\cup\{-\infty,\infty\},\oplus,\infty,+,0)$ is a semiring, where

[TABLE]

Example 1.5*.*

([Gol1999, Example 1.8], [AA1994]) Consider

[TABLE]

where $B(n,i)=\{0,1,2,\cdots,n-1\}$ and

$a\oplus b=a+b$ if $a+b<n$; otherwise, $a\oplus b=c$ is the unique natural number $i\leq c<n$ satisfying $c\equiv a+b$ mod $(n-i);$

$a\odot b=ab$ if $ab<n$; otherwise, $a\odot b=c$ is the unique natural number $i\leq c<n$ with $c\equiv ab$ mod $(n-i).$

Then $B(n,i)$ is a semiring. Notice that $B(n,0)=\mathbb{Z}_{n}$ (a group) and that $B(2,1)=\mathbb{B}$ (the Boolean Algebra).

Example 1.6*.*

([Gol1999, page 150, 154]) Let $S$ be a semiring, $M$ be a left $S$-semimodule and $L\leq_{S}M.$ The **subtractive closure **of $L$ is defined as

[TABLE]

One can easily check that $\overline{L}=Ker(M\overset{\pi_{L}}{\longrightarrow}M/L),$ where $\pi_{L}$ is the canonical projection. We say that $L$ is subtractive (or a $k$-subsemimodule), iff $L=\overline{L}.$ The left $S$-semimodule $M$ is a subtractive semimodule, iff every $S$-subsemimodule $L\leq_{S}M$ is subtractive.

Definition 1.7**.**

[Gol1999, page 71] Let $S$ be a semiring. A subtractive left (right) ideal of $S$ is called a left (right) $k$*-ideal *[Hen1958]. We say that $S$ is a left subtractive (right subtractive) semiring, iff every left (right) ideal of $S$ is subtractive. We say that $S$ is a subtractive semiring, iff $S$ is both left and right subtractive.

Remark 1.8**.**

Whether a left subtractive semiring is necessarily right subtractive was an open problem till a counterexample was given in [KNT2011, Fact 2.1].

Following [BHJK2001], we use the following definitions.

1.9**.**

(cf., [AHS2004]) The category ${}_{S}\mathbf{SM}$ of left semimodules over a semiring $S$ is a variety in the sense of Universal Algebra (closed under homomorphic images, subobjects and arbitrary products). Whence ${}_{S}\mathbf{SM}$ is complete, i.e. has all limits (e.g., direct products, equalizers, kernels, pullbacks, inverse limits) and cocomplete, i.e. has all colimits (e.g., direct coproducts, coequalizers, cokernels, pushouts, direct colimits).

1.10**.**

An $S$-semimodule $N$ is a direct summand of an $S$-semimodule $M$ (i.e. $M=N\oplus N^{\prime}$ for some $S$-subsemimodule $N^{\prime}$ of $M$) if and only if there exists $\alpha\in\mathrm{Comp}(\mathrm{End}(M_{S}))$ s.t. $\alpha(M)=N$ where for any semiring $T$ we set

[TABLE]

Indeed, every direct summand of $M$ is a retract of $M;$ the converse is not true in general. Golan [Gol1999, Proposition 16.6] provided characterizations of direct summands.

Exact Sequences

Throughout, $(S,+,0,\cdot,1)$ is a semiring and, unless otherwise explicitly mentioned, an $S$-module is a *left *$S$-semimodule.

Definition 1.11**.**

A morphism of left $S$-semimodules $f:L\rightarrow M$ is

$k$-normal, iff whenever $f(m)=f(m^{\prime})$ for some $m,m^{\prime}\in M,$ we have $m+k=m^{\prime}+k^{\prime}$ for some $k,k^{\prime}\in Ker(f);$

$i$-normal, iff $im(f)=\overline{f(L)}$ ($:=\{m\in M|\text{ }m+\ell\in L\text{ for some }\ell\in L\}$).

normal, iff $f$ is $k$-normal and $i$-normal.

There are several notions of exactness for sequences of semimodules. In this paper, we use the relatively new notion of exactness introduced by Abuhlail [Abu2014, 2.4] which is stronger than that in the sense of [Tak1982a].

Definition 1.12**.**

([Abu2014, 2.4]) A sequence

[TABLE]

of left $S$-semimodules is exact, iff $f(L)=Ker(g)$ and $g$ is $k$-normal.

1.13**.**

We call a (possibly infinite) sequence of $S$-semimodules

[TABLE]

chain complex, iff $f_{j+1}\circ f_{j}=0$ for every $j;$

exact, iff each partial sequence with three terms $M_{j}\overset{f_{j}}{\rightarrow}M_{j+1}\overset{f_{j+1}}{\rightarrow}M_{j+2}$ is exact.

A short exact sequence (or a Takahashi extension [Tak1982b]) of $S$-semimodules is an exact sequence of the form

[TABLE]

2 Noetherian and Artinian Semirings

As before, $(S,+,0,\cdot,1)$ is a semiring and, unless otherwise explicitly mentioned, an $S$-semimodule is a left $S$-semimodule.

Definition 2.1**.**

A left $S$-semimodule $M$ is

Noetherian (resp., $k$-Noetherian), iff $M$ satisfies the ACC on its $S$-subsemimodules (resp., subtractive $S$-subsemimodules).

Artinian (resp., $k$-Artinian), iff $M$ satisfies the DCC on its $S$-subsemimodules (resp., subtractive $S$-subsemimodules).

The corresponding notions for right $S$-semimodules are defined analogously.

Remark 2.2**.**

Every direct summand of an $S$-semimodule is subtractive. Let $M$ be an $S$-semimodule and $L$ a direct summand of $M.$ Then there exists $N\leq_{S}M$ such that $M=N\oplus L.$ Let $m\in M$ and $\ell,\ell^{\prime}\in L$ be such that $m+\ell=\ell^{\prime}$. Write $m=\widetilde{n}+\widetilde{\ell}$ for some $\tilde{n}\in N$ and $\widetilde{\ell}\in L$, whence $m+\ell=(\tilde{n}+\widetilde{\ell})+\ell=\tilde{n}+(\widetilde{\ell}+\ell)=\ell^{\prime}$. Since the sum $N+L$ is direct, $\tilde{n}=0,$ and thus $m=\widetilde{\ell}\in L$.$\blacksquare$

The following result is an easy observation; however, we highlight it as it will be used frequently in the proofs of the main results.

Lemma 2.3**.**

*Let $M$ be an $S$-semimodule and $N$ a *subtractive $S$-subsemimodules of $M.$ If $M=L\oplus K$ for some $L\leq_{S}N$ and $K\leq_{S}M,$ then

[TABLE]

Proof**.**

Clearly, $L+(K\cap N)\subseteq N$. Let $n\in N.$ Since $M=L+K,$ there exist $k\in K$ and $\ell\in L$ such that $n=\ell+k.$ Since $\ell\in N$ and $N$ is subtractive, we have $k\in N,$ whence $n\in L+(K\cap N)$. So, $N=$ $L+(K\cap N).$ Suppose now that $\ell+k=\ell^{\prime}+k^{\prime}$ for some $\ell,\ell^{\prime}\in L$ and $k,k^{\prime}\in K\cap N$. Since the sum $L+K$ is direct, $\ell=\ell^{\prime}$ and $k=k^{\prime}$.$\blacksquare$

Example 2.4*.*

Let $S:=M_{2}(\mathbb{R}^{+}).$ Consider the left ideals

[TABLE]

and the left ideal

[TABLE]

Then we have $N_{\geq 1}\cap(E_{1}\oplus E_{2})=N_{\geq 1}\cap S=N_{\geq 1},$ while $N_{\geq 1}\cap E1=\{0\}$ and $N_{\geq 1}\cap E_{2}=E_{2}.$ So, we have

[TABLE]

Notice that $N_{\geq 1}\leq_{S}S$ is not subtractive, whence the condition that $N$ is a subtractive subsemimodule of $M$ in Lemma 2.3 cannot be dropped.$\blacksquare$

Definition 2.5**.**

Let $S$ be a semiring, $M$ be a left $S$-semimodule and $N\leq_{S}M.$ A subtractive left $S$-subsemimodule $L\leq_{S}M$ is a maximal subtractive subsemimodule of $N$ if $L\subsetneqq N$ and if $L^{\prime}$ is a subtractive subsemimodule of $M$ with $L\subseteq L^{\prime}\subseteq N$, then $L=L^{\prime}$ or $L^{\prime}=N$.

Lemma 2.6**.**

If $M$ is a $k$-Noetherian left $S$-semimodule, then every non-zero subsemimodule of $M$ contains a maximal subtractive $S$-subsemimodule.

Proof**.**

Let $N\leq_{S}S$ be a non-zero subsemimodule and consider

[TABLE]

Notice that $L_{0}:=\{0_{M}\}\in\mathcal{I}$. If $L_{0}$ is a maximal subtractive subsemimodule of $N,$ then we are done. Otherwise, there exists $L_{1}\in\mathcal{I}$ such that $L_{0}\subsetneqq L_{1}.$ If $L_{1}$ is a maximal subtractive subsemimodule of $M,$ we are done. Otherwise, there exists $L_{2}\in\mathcal{I}$ such that $L_{1}\subsetneqq L_{2}.$ If no such maximal subsemimodule of $N$ exists, we obtain a non-terminating strictly ascending chain

[TABLE]

of $S$-subsemimodules of $N$ which are subtractive subsemimodules of $M$, absurd since $M$ is $k$-Noetherian.$\blacksquare$

Definition 2.7**.**

The semiring $S$ is** left Noetherian** (resp., left $k$-Noetherian), iff ${}_{S}S$ is Noetherian (resp., left $k$-Noetherian), equivalently every ascending chain condition of left (resp., subtractive left) ideals of $S$ terminates;

left Artinian (resp., left $k$-Artinian), iff ${}_{S}S$ is Artinian (resp., left $k$-Artinian), equivalently every descending chain of left (resp., subtractive left) ideals of $S$ terminates.

The **right **($k$-)Noetherian and right ($k$-)Artinian semirings are defined analogously. A semiring which is both left and right ($k$-)Noetherian is called ($k$-)Noetherian, and a semiring which is both left and right ($k$-)Artinian is called ($k$-)Artinian.

Example 2.8*.*

([AD1975]) The semiring $\mathbb{Z}^{+}$ is Noetherian but not Artinian. Setting $I_{k}:=\{0,k,k+1,k+2,\cdots\}$ yields the strictly descending non-terminating chain of ideals of $\mathbb{Z}^{+}:$

[TABLE]

i.e. $\mathbb{Z}^{+}$ is not Artinian.

Lemma 2.9**.**

The only non-trivial proper subtractive left ideals of $S:=M_{2}(\mathbb{R}^{+})$ are

[TABLE]

Proof**.**

We prove this technical lemma is three steps.

Step I:$\ E_{1},$ $E_{2}$ and $N_{r}$ ($r\in\mathbb{R}^{+}\backslash\{0\}$) are subtractive left ideals of $S.$

$E_{1}\leq S$ is a left ideal: for every $a,b,c,d,p,q,r,s\in\mathbb{R}^{+}$ we have

[TABLE]

Moreover, $E_{1}$ is subtractive since

[TABLE]

implies $q=0=s$ and \left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]\in E_{1}. Similarly, $E_{2}$ is a subtractive left ideal of $S$.

For any nonzero $r\in\mathbb{R}^{+}$, $N_{r}$ is a left ideal since (for all $a,b,c,d,k,\ell,m,n\in\mathbb{R}^{+}$) we have

[TABLE]

Moreover, $N_{r}\leq S$ is subtractive since

[TABLE]

whence $c=a+k/r=a+\ell,d=b+m/r=b+n.$ So, $k=r\ell,m=rn$, and \left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]\in N_{r}.

Step II:$\ E_{1},$ $E_{2}$ and $N_{r}$ ($r\in\mathbb{R}^{+}\backslash\{0\}$) are subtractive left ideals of $S.$

Let $I$ be a subtractive left ideal of $M_{2}(\mathbb{R}^{+})$ such that $E_{1}\subsetneqq I$. Then there exists \left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]\in I such that $q\neq 0$ or $s\neq 0$, whence \left[{\begin{array}[]{cc}0&q\\ 0&s\end{array}}\right]\in I as \left[{\begin{array}[]{cc}p&0\\ r&0\end{array}}\right]\in I and

[TABLE]

If $q\neq 0$, then

[TABLE]

If $s\neq 0$, then

[TABLE]

Either way \left[{\begin{array}[]{cc}0&0\\ 0&1\end{array}}\right]\in I, which implies $E_{2}\subseteq I$ and $I=S$. Similarly, if $I$ is a subtractive left ideal of $M_{2}(\mathbb{R}^{+})$ such that $E_{2}\subsetneqq I$, then $I=S$.

Let $r\in\mathbb{R}^{+}\backslash\{0\}$ and $I$ be a subtractive left ideal of $M_{2}(\mathbb{R}^{+})$ such that $N_{r}\subsetneqq I$. Then there exists \left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]\in I such that $k\neq r\ell$ or $m\neq rn$. Without loss of generality, assume that $k<r\ell$. Then $k+p=r\ell$ for some $p\in\mathbb{R}^{+}\backslash\{0\}$. Thus \left[{\begin{array}[]{cc}p&0\\ q&0\end{array}}\right]\in I or \left[{\begin{array}[]{cc}p&0\\ 0&q\end{array}}\right]\in I for some $q\in\mathbb{R}^{+}$ as

[TABLE]

or

[TABLE]

Thus

[TABLE]

or

[TABLE]

Either way we have \left[{\begin{array}[]{cc}1&0\\ 0&0\end{array}}\right]\in I, whence $E_{1}\subsetneqq I$ and $I=S$.

Step III:$\ E_{1},$ $E_{2}$ and $N_{r}$ ($r\in\mathbb{R}^{+}\backslash\{0\}$) are the only subtractive left ideals of $S.$

Let $I$ be a proper non-trivial subtractive left ideal of $S.$ Then \left[{\begin{array}[]{cc}k&\ell\\ m&n\end{array}}\right]\in I\backslash\{0\} for some $k,\ell,m,n\in\mathbb{R}^{+}$. If $k\neq 0$, then

[TABLE]

whence \left[{\begin{array}[]{cc}1&0\\ 0&0\end{array}}\right]\in I, or \left[{\begin{array}[]{cc}k/\ell&1\\ 0&0\end{array}}\right]\in I, and it follows that $I\in\{E_{1},N_{k/\ell},S\}$ as $I$ contains $E_{1}$ or $N_{k/\ell}$. If $\ell\neq 0$, then

[TABLE]

whence \left[{\begin{array}[]{cc}0&0\\ 0&1\end{array}}\right]\in I or \left[{\begin{array}[]{cc}0&0\\ k/\ell&\ell\end{array}}\right]\in I, and so $I\in\{E_{2},N_{k/\ell},S\}$ as $I$ contains $E_{2}$ or $N_{k/\ell}$. If $m\neq 0$, then

[TABLE]

whence \left[{\begin{array}[]{cc}1&0\\ 0&0\end{array}}\right]\in I or \left[{\begin{array}[]{cc}m/n&1\\ 0&0\end{array}}\right]\in I, and it follows that $I\in\{E_{1},N_{m/n},S\}$ as $I$ contains $E_{1}$ or $N_{m/n}$. If $n\neq 0$, then

[TABLE]

whence \left[{\begin{array}[]{cc}0&0\\ 0&1\end{array}}\right]\in I or \left[{\begin{array}[]{cc}0&0\\ m/n&\ell\end{array}}\right]\in I. So, $I\in\{E_{2},N_{m/n},S\}$ as $I$ contains $E_{2}$ or $N_{m/n}$.$\blacksquare$

We provide an example of a semiring which is left $k$-Artinian and left $k$-Noetherian but neither left Artinian (nor left Noetherian):

Example 2.10*.*

Let $S=M_{2}(\mathbb{R}^{+})$. By Lemma 2.9, the only subtractive left ideals of $S$ are [math], $S,$ $E_{1},$ $E_{2}$ and $N_{r}$ ($r\in\mathbb{R}^{+}\backslash\{0\}$). Notice that for $r\neq s$, the left ideals $N_{r},N_{s}$ are not comparable. Thus, the longest ascending (descending) chain of subtractive left ideals of $S$ is $0\subsetneqq N\subsetneqq S$ ($S\varsupsetneqq N\varsupsetneqq 0$) with $N=E_{2}$ or $N=N_{r}$ for some $r\in\mathbb{R}^{+}.$ Whence, $S$ is left $k$-Artinian and left $k$-Noetherian.

On the other hand, for every $r\in\mathbb{R}^{+}$ we have a left ideal of $S$ given by

[TABLE]

Thus, we have an infinite strictly descending chain of left ideal that does not terminate

[TABLE]

i.e. $S$ is not $k$-Artinian. On the other hand, we have an infinite ascending chain of left ideals that does not terminate

[TABLE]

i.e. $S$ is not $k$-Noetherian.$\blacksquare$

An additional example of a $k$-Noetherian semiring that is not Noetherian was communicated to Abuhlail by T. Nam:

Example 2.11*.*

The semiring $\mathbb{R}^{+}[x]$ is $k$-Noetherian but not Noetherian.

Proof**.**

The semiring $\mathbf{B}[x],$ where $\mathbf{B}$ is the Boolean semiring, is not Noetherian. The surjective morphism of semirings

[TABLE]

induces a surjective morphism of semirings $\mathbb{R}^{+}[x]\longrightarrow\mathbf{B}[x],$ whence $\mathbb{R}^{+}[x]$ is not Noetherian.$\blacksquare$

We do not know whether $k$-Artinian semirings are $k$-Noetherian. However, we have the following interesting result.

Lemma 2.12**.**

A left $S$-semimodule $M$ satisfies the ACC on direct summands if and only if $M$ satisfies the DCC on direct summands.

Proof**.**

$(\Longrightarrow)$ Assume that $M$ satisfies the ACC on direct summands. Let

[TABLE]

be a descending chain of direct summands of $M.$ For every $i\in\mathbb{N}$, there exists a direct summand $L_{i}\leq_{S}M$ such that $M=N_{i}\oplus L_{i}.$ Since $M=N_{2}\oplus L_{2}$ and $N_{1}\supseteq N_{2},$ we have by (taking into consideration Remark 2.2):

[TABLE]

Set $K_{1}:=L_{1}$ and $K_{2}:=(N_{1}\cap L_{2})\oplus L_{1}$, so that $N_{1}\oplus K_{1}=M=N_{2}\oplus K_{2}$ and $K_{1}\subseteq K_{2}.$

Now, $N_{2}\supseteq N_{3}$ and $M=N_{3}\oplus L_{3},$ whence $N_{2}\overset{\text{Lemma \ref{lemint}}}{=}N_{3}\oplus(N_{2}\cap L_{3})$ and so

[TABLE]

Set $K_{3}:=(N_{2}\cap L_{3})\oplus K_{2}$, so that $M=N_{3}\oplus K_{3}$ and $K_{2}\subseteq K_{3}$. Continuing this way, we obtain an ascending chain

[TABLE]

of direct summands of ${}_{S}M.$ By our assumption, the ascending chain (8) terminates, whence there exists $t\in\mathbb{N}$ such that $K_{i}=K_{t}$ for any $i\geq t$. For any $i\geq t,$ we have $N_{t}\supseteq N_{i},$ $M=N_{i}\oplus K_{i}$ and $N_{t}\cap K_{t}=0$ and so

[TABLE]

thus the descending chain (7) terminates.

$(\Longleftarrow)$ Assume that $M$ satisfies the DCC on direct summands. Let

[TABLE]

be an ascending chain of direct summands of $M$. For every $i\in\mathbb{N}$, there exists an $S$-subsemimodule $N_{i}\leq_{S}M$ such that $M=L_{i}\oplus N_{i};$ in particular $M=L_{1}\oplus N_{1}.$ Since $L_{1}\subseteq L_{2}$ it follows (taking into consideration Remark 2.2) that $L_{2}\overset{\text{Lemma \ref{lemint}}}{=}L_{1}\oplus(L_{2}\cap N_{1}),$ whence

[TABLE]

Since $L_{2}\cap N_{1}\subseteq N_{1}$ it follows that

[TABLE]

whence

[TABLE]

Setting $N_{1}^{\prime}:=N_{1}$ and $N_{2}^{\prime}:=N_{1}\cap(L_{1}\oplus N_{2})$, we have $L_{1}\oplus N_{1}^{\prime}=M=L_{2}\oplus N_{2}^{\prime}$ where $N_{1}^{\prime}\supseteq N_{2}^{\prime}$. Since $M=L_{2}\oplus N_{2}^{\prime}$ and $L_{2}\subseteq L_{3},$ it follows that $L_{3}\overset{\text{Lemma \ref{lemint}}}{=}L_{2}\oplus(L_{3}\cap N_{2}^{\prime}),$ whence

[TABLE]

Since $L_{3}\cap N_{2}^{\prime}\subseteq N_{2}^{\prime},$ we have

[TABLE]

Setting $N_{3}^{\prime}:=N_{2}^{\prime}\cap(L_{2}\oplus N_{3}),$ we have $N_{2}^{\prime}\supseteq N_{3}^{\prime}$ and

[TABLE]

Continuing this process, we obtain a descending chain

[TABLE]

of *direct summands *of $M$ such that $M=L_{i}\oplus N_{i}^{\prime}$ for every $i\in\mathbb{N}$. By our assumption, the descending chain (10) terminates, *i.e. *there exists some $k\in\mathbb{N}$ such that $N_{i}^{\prime}=N_{k}^{\prime}$ for every $i\geq k$.

Now, for every $i\geq k$, we have $L_{k}\subseteq L_{i},$ $M=L_{k}\oplus N_{k}^{\prime}$ and $L_{i}\cap N_{i}^{\prime}=0$ and so

[TABLE]

Thus the ascending chain (9) terminates.$\blacksquare$

A ring in which every left ideal is a direct summand is left Artinian and left Noetherian [Wis1991, 3.4 and 4.1] (in fact, left semisimple). The following result extends this fact to semirings.

Theorem 2.13**.**

If every subtractive left ideal of $S$ is a direct summand, then $S$ is left $k$-Artinian and left $k$-Noetherian.

Proof**.**

Assume that every subtractive left ideal of $S$ is a direct summand.

Claim I: $S$ is left $k$-Artinian.

Suppose that

[TABLE]

is a strictly descending chain of left subtractive ideals of $S$ that does not terminate. For every $k\in\mathbb{N},$ there exists, by our assumption, some left ideal $N_{k}\leq_{S}S$ such that $S=I_{k}\oplus N_{k}.$ The left ideals $I_{k},N_{k}$ are non-zero as the chain does not terminate, and are subtractive by Remark 2.2.

Since $I_{1}\supseteq I_{2}$ and $S=I_{2}\oplus N_{2}$, we have

[TABLE]

Then $J_{1}:=I_{1}\cap N_{2}$ is a subtractive left ideal of $S,$ which is non-zero as $I_{1}\supsetneqq I_{2},$ and $I_{1}=I_{2}\oplus J_{1}.$ Since $I_{2}\supseteq I_{3}$ and $S=I_{3}\oplus N_{3}$, we have

[TABLE]

Then $I_{2}\cap N_{3}$ is a subtractive left ideal of $S,$ which is non-zero as $I_{2}\supsetneqq I_{3},$ and $I_{1}=I_{2}\oplus J_{1}=I_{3}\oplus J_{2}\oplus J_{1}.$ Continuing this process, we obtain at the $k$th step, a non-zero subtractive left ideal $J_{k}\leq_{S}S$ such that

[TABLE]

Setting $J_{i}^{\prime}:=J_{1}\oplus\cdots\oplus J_{i}$ for each $i\in\mathbb{N}$, we have $S=J_{i}^{\prime}\oplus I_{i+1}\oplus N_{1}$ whence $J_{i}^{\prime}$ is subtractive (by Remark 2.2). One can easily show that $J:=\bigcup\limits_{i\in\mathbb{N}}J_{i}^{\prime}$ is subtractive.

By our assumption, $S=J\oplus N$ for some left ideal of $N\leq_{S}S$. Thus $1_{S}=j+n$ for some $j\in J$ and $n\in N$. Since $j\in J_{i}^{\prime}$ for some $i\in\mathbb{N}$, it can be written in a unique way as $j=j_{1}+j_{2}+...+j_{i}$ for some uniquely determined $j_{k}\in J_{k},$ $k=1,2,...,i$. Since $J_{i+1}^{\prime}\subseteq J$, the sum $J_{i+1}^{\prime}+N$ is direct, whence the sum $J_{1}+J_{2}+...+J_{i}+J_{i+1}+N$ is direct. Setting

[TABLE]

this means that the sum $J_{i+1}+K$ is direct. For any $s_{i+1}\in J_{i+1}\backslash\{0\},$ we have

[TABLE]

where $s_{i+1}j_{k}\in J_{k}$ for $k=1,2,...,i$ and $s_{i+1}n\in N.$ It follows that $s_{i+1}\in J_{i+1}\cap K=0$, absurd since $s_{i+1}\neq 0$. So, the descending chain (11) terminates.

Claim II: $S$ is left $k$-Noetherian.

Let

[TABLE]

be an ascending chain of subtractive left ideals of $S$. Since every direct summand of ${}_{S}S$ is subtractive (by Remark 2.2), it follows from the proof of Claim I that ${}_{S}S$ satisfies DCC on direct summands, whence ${}_{S}S$ satisfies ACC on direct summands by Lemma 2.12. Since (12) is an ascending chain of subtractive left ideals of $S,$ whence of direct summands of ${}_{S}S$ (by our assumption), the chain terminates.$\blacksquare$

Example 2.14*.*

Let $p$ be a prime number. Every subtractive ideal of the semiring $S=B(p+1,p)$ is a direct summand, and $S$ is $k$-Artinian and $k$-Noetherian.

Proof**.**

$S$ has no non-trivial subtractive ideals, thus every subtractive left ideal of $S$ is a direct summand. Notice that $S$ is $k$-Artinian and $k$-Noetherian since it has finitely many elements.$\blacksquare$

Example 2.15*.*

Let $S:=\mathbb{B}^{\mathbb{N}}$ with the canonical structure of a semiring induced by that on $\mathbb{B}$. Then $S$ has a subtractive left ideal which is not a direct summand and $S$ is neither $k$-Artinian nor $k$-Noetherian.

Proof**.**

The subtractive left ideal $\bigoplus\limits_{n\in\mathbb{N}}\mathbb{B}$ is not a direct summand. Notice that neither the ascending chain

[TABLE]

nor the descending chain

[TABLE]

terminates, thus $S$ is neither $k$-Noetherian nor $k$-Artinian.$\blacksquare$

2.16**.**

Let $I$ be a left $S$-semimodule.

For a left $S$-semimodule $M,$ we say that $I$ is

$M$-injective [Gol1999, page 197], iff for every injective $S$-linear map $f:L\rightarrow M$ and any $S$-linear map $g:L\rightarrow I,$ there exists an $S$-linear map $h:M\rightarrow I$ such that $h\circ f=g;$

[TABLE]

$M$-$i$-injective [Alt2003], iff for every normal monomorphism $f:L\rightarrow M$ and any $S$-linear map $g:L\rightarrow I,$ there exists an $S$-linear map $h:M\rightarrow I$ such that $h\circ f=g;$

[TABLE]

$M$-$e$-injective [AIKN2018], iff for every short exact sequence $0\longrightarrow L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N\longrightarrow 0$ of left $S$-semimodules, the following sequence $0\longrightarrow Hom_{S}(N,I)\overset{(g,I)}{\longrightarrow}Hom_{S}(M,I)\overset{(f,I)}{\longrightarrow}Hom_{S}(L,I)\longrightarrow 0$ of commutative monoids is exact.

We say that $I$ is injective (resp. $i$-injective, $e$-injective), iff $I$ is $M$-injective (resp. $M$-$i$-injective, $M$-$e$-injective) for every left $S$-semimodule $M.$

A Bass-Papp Theorem for Semirings

The celebrated Bass-Papp Theorem states that a ring $R$ is left (right) Noetherian if and only if every direct sum of left (right) injective $R$-modules is ($R$-)injective (e.g., [Rot2009, 3.39], [Gri2007, page 407]). This characterization was extended to semirings: $S$ by Il’in and

An $e$-injective version of this theorem was obtained by Abuhlail et al.:

Proposition 2.17**.**

*([Ili2010, Theorem 3.6] ,[AIKN2018, Theorem 5.5]) *The following are equivalent for a semiring $S:$

1. (1)

$S$* is a left Noetherian ring;* 2. (2)

every direct sum of injective left $S$-semimodules is injective and the left $S$-semimodule $S/V(S)$ can be embedded in an injective left $S$-semimodule. 3. (3)

every direct sum of $e$-injective left $S$-semimodules is $e$-injective and the left $S$-semimodule $S/V(S)$ can be embedded in an $e$-injective left $S$-semimodule.

Notice that the assumptions in Proposition 2.17 force the semiring $S$ to be a ring. In what follows we provide a partial version of the Bass-Papp characterization for left $k$-Noetherian semirings.

2.18**.**

We say that a semiring $S$ has enough $S$-$i$-injective left semimodules, iff every left $S$-semimodule can be embedded into an $S$-$i$-injective left $S$-semimodule.

Theorem 2.19**.**

Let $S$ be a semiring with enough left $S$-$i$-injective left semimodules. If every direct sum of left $S$-$i$-injective $S$-semimodules is $S$-$i$-injective, then $S$ is left $k$-Noetherian.

Proof**.**

Let

[TABLE]

be a chain of subtractive left ideals of $S$ and consider the left ideal $L=:\bigcup\limits_{n\in\mathbb{N}}L_{n}.$ It is clear that $L$ is subtractive. By our assumption, there exists for every $n\in\mathbb{N}$ an $S$-$i$-injective left $S$-semimodule $J_{n}$ and an embedding $S/L_{n}\overset{\iota_{n}}{\hookrightarrow}J_{n}.$ Set $J:=\bigoplus\limits_{n\in\mathbb{N}}J_{n}$ and consider the $S$-linear map

[TABLE]

Notice that $\varphi$ is well defined as each $x\in L$ belongs to $L_{n}$ for some $n\in\mathbb{N}$ and so $\varphi_{k}(x)=0$ for all $k\geq n,$ i.e. $\varphi(x)=\sum\limits_{k=1}^{\infty}\varphi_{k}(x)=\sum\limits_{k=1}^{n-1}\varphi_{k}(x)$.

By our assumption, $J$ is $S$-$i$-injective and so there exists an $S$-linear map $\psi:S\longrightarrow J$ such that $\psi\circ\iota=\varphi.$

[TABLE]

Let $\psi(1_{S})=\sum t_{k}\in J.$ Then $\psi(1_{S})\in\bigoplus\limits_{k=1}^{m-1}J_{k}$ for some $m,$ whence $\psi(x)=\psi(x\cdot 1_{S})=x\psi(1_{S})\in\bigoplus\limits_{k=1}^{m-1}J_{k}$ for every $x\in L.$ In particular, $\varphi_{m}(x)=(\pi_{m}\circ\phi)(x)=0$ where $\pi_{m}$ is the projection on $J_{m}$. Thus $x\in L_{n}$ and $L=L_{n},$ whence $L_{k}=L_{n}$ for all $k\geq n,$ i.e. the chain terminates. Consequently, $S$ is $k$-Noetherian.$\blacksquare$

Example 2.20*.*

If $S$ is an additively idempotent semiring such that every direct sum of left $S$-$i$-injective $S$-semimodules is $S$-$i$-injective, then $S$ is left $k$-Noetherian. This results from Theorem 2.19 and the fact that every semimodule over an additively idempotent semiring can be embedded into an $e$-injective (whence an $i$-injective) left $S$-semimodule [AIKN2018, 4.5].

Theorem 2.21**.**

If $S$ is a semiring such that every short exact sequence of left $S$-semimodules $0\rightarrow L\rightarrow S\rightarrow N\rightarrow 0$ is left splitting, then $S$ is a left $k$-Noetherian.

Proof**.**

Let

[TABLE]

be a non-terminating ascending chain of subtractive left ideals of $S$. Notice that $N:=\bigcup\limits_{i\in\mathbb{N}}N_{i}$ is a subtractive ideal of $S$, whence (by assumption) the following short exact sequence

[TABLE]

of left $S$-semimodules is left splitting. Let $h:S\rightarrow N$ be an $S$-linear map such that $h\circ\iota=id_{N}$. Then $h(1_{S})\in N$, that is $h(1_{S})\in N_{i}$ for some $i\in\mathbb{N}$. If $x\in N_{i+1}\backslash N_{i},$ then

[TABLE]

a contradiction. Hence $S$ is $k$-Noetherian.$\blacksquare$

Corollary 2.22**.**

If $S$ is a semiring such that every subtractive left ideal is $S$-$i$-injective, then $S$ is a left $k$-Noetherian semiring.