On Semisimple Semirings

TL;DR

This paper explores the properties of ideal-semisimple and congruence-semisimple semirings, providing new characterizations and extending classical ring results to broader semiring classes.

Contribution

It introduces novel characterizations of semisimple semirings using e-projective and e-injective semimodules, extending known ring results to non-subtractive commutative semirings.

Findings

New characterizations of semisimple semirings

Extension of ring semisimplicity results to semirings

Broader understanding of semiring structures

Abstract

We investigate ideal-semisimple and congruence-semisimple semirings. We give several new characterizations of such semirings using e-projective and e-injective semimodules. We extend several characterizations of semisimple rings to (not necessarily subtractive) commutative semirings.

Equations

$$a\cdot 0$$

$$a(b+c)$$

$$V(S):=\{s\in S\mid s+t=0\text{ for some }t\in S\}.$$

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

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

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

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

$$m\rho m^{\prime }\text{ and }n\rho n^{\prime }⟹(m+m^{\prime })\rho (n+n^{\prime }),$$

$$m\rho m^{\prime }⟹(sm)\rho (s{m}^{\prime }).$$

$$s\rho t\text{ and }{s}^{\prime }\rho t^{\prime }⟹(s+s^{\prime })\rho (t+t^{\prime })\text{ and }(s{s}^{\prime })\rho (t{t}^{\prime }).$$

$$m{\equiv }_N{m}^{\prime }\iff m+n=m^{\prime }+n^{\prime }\text{ for some }n,n^{\prime }\in N.$$

$${\Delta }_M:=\{(m,m)\mid m\in M\}$$

$$\rho =\{(a,b)|a,b\in D\backslash \{0\}\}\cup \{(0,0)\}$$

$$m{\equiv }_f{m}^{\prime }⟺f(m)=f(m^{\prime }).$$

$$\sum _{a\in A}{l}_a=\sum _{a\in A}{l}_a^{\prime }⟹l_a=l_a^{\prime }\text{ for all }a\in A.$$

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

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

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

$$L\stackrel{f}{⟶}M\stackrel{g}{⟶}N$$

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

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

$$Ho{m}_S(P,-):{}_S\mathbf{SM}⟶{}_{{\mathbb{Z}}^{+}}\mathbf{SM}$$

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

$$0⟶Ho{m}_S(P,L)\stackrel{(P,f)}{⟶}Ho{m}_S(P,M)\stackrel{(P,g)}{⟶}Ho{m}_S(P,N)⟶0.$$

$$Ho{m}_S(-,J):{}_S\mathbf{SM}⟶{}_{{\mathbb{Z}}^{+}}\mathbf{SM}$$

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

$$0⟶Ho{m}_S(N,J)⟶Ho{m}_S(M,J)⟶Ho{m}_S(L,J)⟶0.$$

$$0\to A\stackrel{f}{⟶}B\stackrel{g}{⟶}C\to 0$$

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

$${\mathcal{D}}_1:=\{D⫌D_0|D\text{ is a direct summand of }S\}$$

$${\mathcal{D}}_{-1}:=\{D⫋D_0|D\text{ is a direct summand of }S\}$$

Full text

Introduction

Semirings (defined, roughly, as rings not necessarily with subtraction) can be considered as a generalization of both rings and distributive bounded lattices. Semirings, and their semimodules (defined, roughly, as modules not necessarily with subtraction), have wide applications in many aspects of Computer Science and Mathematics, e.g., Automata Theory [HW1998], Tropical Geometry [Gla2002] and Idempotent Analysis [LM2005]. Many of these applications can be found in Golan’s book [Gol1999], which is considered a main reference in this topic.

Several papers by Abuhlail, I’llin, Katsov and Nam (among others) prepared the stage for a homological characterization of special classes of semirings using special classes of projective, injective and flat semimodules (cf., [KNT2009], [Ili2010], [KN2011], [Abu2014], [KNZ2014], [AIKN2015], [IKN2017], [AIKN2018]).

The notions of projective and injective objects can be defined in any category relative to a suitable factorization system of its arrows. Projective, injective and flat semimodules have been studied intensively (see [Gla2002] for details). Recently, left (right) $V$-semirings, all of whose congruence-simple left (right) semimodules are injective have been completely characterized in [AIKN2015], and ideal-semisimple semirings all of whose left cyclic semimodules are projective have been investigated in [IKN2017].

In addition to the categorical notions of projective and injective semimodules over a semiring, new notions of projectivity and injectivity of semimodules over semirings were considered by the first author who introduced the so called $e$-projective and $e$-injective semimodules [Abu2014-CA]. One reason for the interest in such notions is the phenomenon that assuming that all semimodules of a given semiring $S$ are projective (injective) forces the semiring to be a ring (cf., [Ili2010, Theorem 3.4]).

The paper is divided into two sections.

In Section 1, we collect the basic definitions, examples and preliminaries used in this paper. Among others, we include the definitions and basic properties of *exact sequences *as defined by Abuhlail [Abu2014].

In Section 2, we investigate ideal-semisimple and congruence-semisimple semirings. A semiring $S$ is left (right) ideal-semisimple, iff $S$ is a direct sum of ideal-simple left (right) ideals. By [HW1996, Theorem 7.8], $S$ is left (right) ideal-semisimple if and only if $S\simeq M_{n_{1}}(D_{1})\times\cdots\times M_{n_{k}}(D_{k})$, where $D_{i}$ is a division semiring and $M_{n_{i}}(D_{i})$ is the semiring of $n_{i}\times n_{i}$-matrices over $D_{i}$ for each $i=1,\cdots,k.$ A left subtractive semiring $S$ was shown to be left ideal-semisimple if and only if every left $S$-semimodule is $S$-$k$-projective ($S$-$k$-injective) [KNT2009, Theorem 4.4]. In Proposition 2.19 (Proposition 2.21), we show that a semiring $S$ over which every left $S$-semimodule is $S$-$k$-projective ($S$-$k$-injective) is a finite direct sum of irreducible summands.

In Section 3, we restrict our attention to commutative semirings. In Theorem 3.11, we extend several classical characterizations of semisimple semirings to commutative, not necessarily subtractive, semirings. In Theorem 3.13, we show that a commutative semiring $S$ is ideal-semisimple if and only if every $S$-semimodule is $S$-$e$-injective ($S$-$k$-injective) and $S$ satisfies some technical condition. The two results are combined in Theorem 3.12 to provide a complete characterization of commutative ideal-semisimple semirings. The congruence-semisimple version of this main result is given in Theorem 3.14. Examples 3.10 and 3.15 demonstrate that the conditions assumed in our main results in this section, in particular the commutativity of the base semiring, cannot be dropped.

1 Preliminaries

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

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). We write $L\leq_{S}M$ to indicate that $L$ is a subsemimodule of the left (right) $S$-semimodule $M.$

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. If moreover, $(S\backslash\{0\},\cdot,1)$ is a group, we say that $S$ is a semifield.

• •

Let

[TABLE]

If $V(S)=\{0\}$, we say that $S$ is zerosumfree. Notice that $V(S)=S$ if and only if $S$ is a ring.

• •

The set of *cancellative elements of *$S$ is defined as

[TABLE]

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

Examples 1.4*.*

([Gol1999])

• •

Every ring is a cancellative semiring.

• •

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

• •

$(\mathbb{Z}^{+},+,0,\cdot,1)$ (resp. $(\mathbb{Q}^{+},+,0,\cdot,1),$ $(\mathbb{R}^{+},+,0,\cdot,1)$), the set of non-negative integers (resp. non-negative rational numbers, non-negative real numbers) is a commutative cancellative 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 semiring called the Boolean semiring. $\mathbb{B}$ is an semifield which is not a field.

• •

The max-plus algebra $\mathbb{R}_{\max,+}:=(\mathbb{R}\cup\{-\infty\},\max,-\infty,+,0)$ is an additively idempotent 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 $B(n,i)=\{0,1,2,...,n-1\}$ and

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

$a\odot b=ab$ if $ab<n$; otherwise, $a\odot b=c$ with $i\leq c<n$ is the unique natural number 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 $L\subseteq M.$ The **subtractive closure **of $L$ is defined as

[TABLE]

One can easily check that $\overline{L}=Ker(M\overset{\pi}{\longrightarrow}M/L),$ where $\pi$ is the canonical projection. We say that $L$ is subtractive, 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. We say that $S$ is a **left subtractive semiring **(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).

Semisimple Semimodules

1.10**.**

[Gol1999, page 162] Let $S$ be a semiring.

An equivalence relation $\rho$ on a left $S$-semimodule $M$ is a congruence relation, iff it preserves the addition and the scalar multiplication on $M,$ *i.e. *for all $s\in S$ and $m,m^{\prime},n,n^{\prime}\in M:$

[TABLE]

[TABLE]

A congruence relation on the semiring $S$ is an equivalence relation $\rho$ on $S$ such that or all $s,s^{\prime},t,t^{\prime}\in S:$

[TABLE]

Example 1.11*.*

Let $S$ be a semiring, $M$ a left $S$-semimodule and $N\leq_{S}M.$ The Bourne relation $\equiv_{N}$ on $M$ is defined as:

[TABLE]

It is clear that $\equiv_{N}$ is a congruence relation. Moreover, $M/N=M/\equiv_{N}=\{[m]_{N}$ $|$ $m\in M\}$ $(=M/\overline{N})$ is a left $S$-semimodule, the canonical surjective map $\pi_{N}:M\longrightarrow M/N$ is $S$-linear, and $Ker(\pi_{N})=\overline{N}.$ In particular, $Ker(\pi_{N})=0$ if and only if $N\leq_{S}M$ is subtractive (this explains why subtractive ideals are called $k$-ideals in many references).

Following [BHJK2001], we use the following definitions.

Definition 1.12**.**

Let $S$ be a semiring. A left $S$-semimodule $M$ is

ideal-simple, iff 0 and $M$ are the only $S$-subsemimodules of $M$;

congruence-simple, iff

[TABLE]

and $M\times M$ are the only congruence relations on $M$.

Definition 1.13**.**

We say that the semiring $S$ is

**left ideal-simple **(right ideal-simple), iff [math] and $S$ are the only left (right) ideals of $S;$ equivalently, $S$ is ideal-simple as a left (right) $S$-semimodule.;

left congruence-simple (right congruence-simple), iff $S$ is congruence-simple as a left (right) $S$-semimodule;

ideal-simple, iff [math] and $S$ are the only (two-sided) ideals of $S;$

congruence-simple, iff $\Delta_{S}$ and $S\times S$ are the only congruence relations on the semiring $S.$

Remark 1.14**.**

If $M$ is a congruence-simple left $S$-semimodule, then the only subtractive $S$-subsemimodules of $M$ are [math] and $M.$ To show this, suppose that $N\neq 0$ is a subtractive $S$-subsemimodule of $M.$ Then $\equiv_{N}$ is a congruence relation on $M$ with $n\equiv_{N}0$ for some $n\in N\backslash 0$. Thus $\equiv_{N}\neq\Delta_{M}$, which implies $\equiv_{N}=M^{2}$ as $M$ is congruence-simple. If $m\in M$, then $mM^{2}0$, that is $m\equiv_{N}0$. Therefore, there exist $n,n^{\prime}\in N$ such that $m+n=n^{\prime}$. Since $N$ subtractive, $m\in N$. Hence $M=N$.

Example 1.15*.*

[KNZ2014, 3.7 (b)] Let $(M,+,0)$ be a finite lattice that is not distributive. The endomorphism semiring $\mathbf{E}_{M}$ of $M$ is a congruence-simple semiring which is not ideal-simple.

Example 1.16*.*

(cf., [KNZ2014, 3.7 (c)]) Every zerosumfree division semiring $D$ that is not isomorphic to $\mathbb{B}$ (e.g., $\mathbb{R}^{+}$) is an ideal-simple semiring which is not congruence-simple since

[TABLE]

is a non-trivial non-universal congruence relation on $D.$

Lemma 1.17**.**

A left $S$-semimodule $M$ is congruence-simple if and only if every non-zero $S$-linear map from $M$ is injective.

Proof**.**

($\Rightarrow$) Let $f:M\rightarrow N$ be a non-zero $S$-linear map and consider the congruence relation on $M$ defined by

[TABLE]

Pick some $m\in M\backslash\{0\}$ such that $f(m)\neq 0$. Since $\equiv_{f}$ is a congruence relation on $M$ with $m\not\equiv_{f}0$, we know $\equiv_{f}\neq M^{2}.$ It follows that $\equiv_{f}=\Delta_{M}$ as $M$ is congruence-simple. Hence $f$ is injective.

($\Leftarrow$) Assume that $M$ is congruence-simple. Let $\rho$ be a congruence relation on $M$. The canonical map $f:M\rightarrow M/\rho$ is $S$-linear. If $f=0$, then $[m]_{\rho}=[0]_{\rho}$ for every $m\in M$, that is $m\rho 0$ for every $m\in M$ and $m\rho m^{\prime}$ for every $m,m^{\prime}\in M$. If $f\neq 0$, then $f$ is injective, that is $[m]_{\rho}\neq[m^{\prime}]_{\rho}$ whenever $m\neq m^{\prime}$. Thus $m\rho m^{\prime}$ whenever $m\neq m^{\prime}$ and $\rho=\Delta_{M}$.$\blacksquare$

Lemma 1.18**.**

A left $S$-semimodule $M$ is ideal-simple if and only if every non-zero $S$-linear map to $M$ is surjective.

Proof**.**

($\Rightarrow$) Let $f:L\rightarrow M$ be a non-zero $S$-linear map. Then there exists $l\in L\backslash 0$ such that $f(l)\neq 0$. Thus, $f(L)$ is a non-zero subsemimodule of $M$ and so $f(L)=M$ as $M$ ideal-simple.

($\Leftarrow$) Let $K$ be a subsemimodule of $M$. Then the embedding $f:K\rightarrow M$ is an $S$-linear map. If $f=0$, then $K=f(K)=0$. If $f\neq 0$, then $f$ is surjective, that is $K=f(K)=M$.$\blacksquare$

1.19**.**

([Gol1999, page 184]) Let $S$ be a semiring. A left $S$-semimodule $M$ is the direct sum of a family $\{L_{\lambda}\}_{\lambda\in\Lambda}$ of $S$-subsemimodules $L_{\lambda}\leq_{S}M,$ and we write $M=\bigoplus\limits_{\lambda\in\Lambda}L_{\lambda}$, iff every $m\in M$ can be written in a unique way as a finite sum $m=l_{\lambda_{1}}+\cdots+l_{\lambda_{k}}$ where $l_{\lambda_{i}}\in L_{\lambda_{i}}$ for each $i=1,\cdots,k.$ Equivalently, $M=\bigoplus\limits_{\lambda\in\Lambda}L_{\lambda}$ if $M=\sum\limits_{\lambda\in\Lambda}L_{\lambda}$ and for each finite subset $A\subseteq\Lambda$ with $l_{a},l_{a}^{\prime}\in L_{a},$ we have:

[TABLE]

1.20**.**

An $S$-semimodule $N$ is a retract of an $S$-semimodule $M,$ iff there exists a (surjective) $S$-linear map $\theta:M\longrightarrow N$ and an (injective) $S$-linear map $\psi:N\longrightarrow M$ such that $\theta\circ\psi=\mathrm{id}_{N}$ (equivalently, $N\simeq\alpha(M)$ for some idempotent endomorphism $\alpha\in\mathrm{End}(M_{S})$).

1.21**.**

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; for example N_{1}=\left\{\left[{\begin{array}[]{cc}a&a\\ b&b\end{array}}\right]|\text{ }a,b\in\mathbb{R}^{+}\right\} is a retract of $M_{2}(\mathbb{R^{+}})$ that is not a direct summand. Golan [Gol1999, Proposition 16.6] provided characterizations of direct summands.

Remarks 1.22**.**

Let $M$ be a left $S$-semimodule and $K,L\leq_{S}M$ be $S$-semimodules of $M.$

1. (1)

If $K+L$ is direct, then $K\cap L=0.$ The converse is not true in general. 2. (2)

If $M=K\oplus L$, then $M/K\simeq L.$

Example 1.23*.*

Let $S=M_{2}(\mathbb{R}^{+})$. Notice that

[TABLE]

are left ideals of $S$ with $E_{1}\cap N_{\geq 1}=\{0\}$. However, the sum $E_{1}+N_{\geq 1}$ is not direct since

[TABLE]

Lemma 1.24**.**

([Gol1999, Proposition 16.6]) Let $S$ be a semiring. The following are equivalent for a left $S$-semimodule $M$ and $N\leq_{S}M:$

1. (1)

$N$* is a direct summand of $M$ (i.e. $M=N\oplus N^{\prime}$ for some $S$-subsemimodule $N^{\prime}$ of $M$);* 2. (2)

$N=\alpha(M)$* for some $\alpha\in\mathrm{Comp}(\mathrm{End}(M_{S}));$* 3. (3)

$\exists$* $N^{\prime}\leq_{S}M$ such that $M=N+N^{\prime}$ and the restrictions to $\equiv_{N}$ to $N^{\prime}$ and the restriction $\equiv_{N^{\prime}}$ to $N$ are trivial.*

Exact Sequences

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

Definition 1.25**.**

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 $\mathop{\mathrm{I}m}(f)=\overline{f(L)}$ ($:=\{m\in M|\text{ }m+l\in L\text{ for some }l\in L\}$).

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

There are several notions of exactness for sequences of semimodules. In this paper, we use the relatively new notion introduced by Abuhlail:

Definition 1.26**.**

([Abu2014, 2.4]) A sequence

[TABLE]

of left $S$-semimodules is

exact, iff $f(L)=Ker(g)$ and $g$ is $k$-normal;

semi-exact, iff $\overline{f(L)}=Ker(g);$

proper-exact, iff $f(L)=Ker(g).$

1.27**.**

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

[TABLE]

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

exact (resp., proper-exact, semi-exact, quasi-exact) if 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 (resp., proper-exact, semi-exact, quasi-exact).

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

[TABLE]

The following result shows some of the advantages of the Abuhlail’s definition of exact sequences over the previous ones:

Lemma 1.28**.**

Let $L,M$ and $N$ be $S$-semimodules.

1. (1)

$0\longrightarrow L\overset{f}{\longrightarrow}M$* is exact if and only if $f$ is injective.* 2. (2)

$M\overset{g}{\longrightarrow}N\longrightarrow 0$* is exact if and only if $g$ is surjective.* 3. (3)

$0\longrightarrow L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N$* is semi-exact and $f$ is normal (proper-exact and $f$ is normal) if and only if $L\simeq\mathrm{Ker}(g).$* 4. (4)

$0\longrightarrow L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N$* is exact if and only if $L\simeq\mathrm{Ker}(g)$ and $g$ is $k$-normal.* 5. (5)

$L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N\longrightarrow 0$* is semi-exact and $g$ is normal if and only if $N\simeq M/f(L).$* 6. (6)

$L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N\longrightarrow 0$* is exact if and only if $N\simeq M/f(L)$ and $f$ is $i$-normal.* 7. (7)

$0\longrightarrow L\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}N\longrightarrow 0$* is exact if and only if $L\simeq\mathrm{Ker}(g)$ and $N\simeq M/L.$*

Corollary 1.29**.**

The following assertions are equivalent:

1. (1)

$0\rightarrow L\overset{f}{\rightarrow}M\overset{g}{\rightarrow}N\rightarrow 0$* is an exact sequence of $S$-semimodules;* 2. (2)

$L\simeq\mathrm{Ker}(g)$* and $N\simeq M/f(L)$;* 3. (3)

$f$* is injective, $f(L)=\mathrm{Ker}(g),$ $g$ is surjective and ($k$-)normal.*

In this case, $f$ and $g$ are normal morphisms.

2 Semisimple Semirings

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

Definition 2.1**.**

Let $S$ be a semiring. A left $S$-semimodule $M$ is called

ideal-semisimple, iff $M=\mathop{\displaystyle\bigoplus}\limits_{\lambda\in\Lambda}M_{\lambda},$ a direct sum of ideal-simple $S$-subsemimodules;

congruence-semisimple, iff $M=\mathop{\displaystyle\bigoplus}\limits_{\lambda\in\Lambda}M_{\lambda},$ a direct sum of congruence-simple $S$-subsemimodules.

Definition 2.2**.**

A semiring $S$ is

left ideal-semisimple (resp., right ideal-semisimple), iff $S$ is ideal-semisimple as a left (right) $S$-semimodule, equivalently $S$ is a finite direct sum of ideal-simple left (right) ideals.

left congruence-semisimple (resp., right congruence-semisimple), iff $S$ is congruence-semisimple as a left (right) $S$-semimodule; equivalently $S$ is a finite direct sum of congruence-simple left (right) ideals.

Definition 2.3**.**

([AIKN2018]) A left $S$-semimodules $P$ is

$M$-$e$-projective (where $M$ is a left $S$-semimodule), iff the covariant functor

[TABLE]

transfers every short exact sequence of left $S$-semimodules

[TABLE]

into a short exact sequence of commutative monoids

[TABLE]

We say that $P$ is $e$-projective, iff $P$ is $M$-$e$-projective for every left $S$-semimodule $M$.

2.4**.**

Let $P$ be a left $S$-semimodule. For a left $S$-semimodule $M,$ we say that $P$ is $M$-projective (resp. $M$-$k$-projective), iff for every (normal) surjective $S$-linear map $f:M\rightarrow N$ and an $S$-linear map $g:P\rightarrow N,$ there exists an $S$-linear map $h:P\rightarrow M$ such that $f\circ h=g.$

We say that $P$ is projective (resp., $k$-projective), iff $P$ is $M$-projective (resp., $M$-$k$-projective) for every left $S$-semimodule $M.$

Definition 2.5**.**

([Abu2014-CA, 1.24]) Let $M$ be a left $S$-semimodule. A left $S$-semimodules $J$ is $M$-$e$-injective, iff the contravariant functor

[TABLE]

transfers every short exact sequence of left $S$-semimodules

[TABLE]

into a short exact sequence of commutative monoids

[TABLE]

We say that $J$ is $e$-injective, iff $J$ is $M$-$e$-injective for every left $S$-semimodule $M.$

2.6**.**

Let $I$ be a left $S$-semimodule. For a left $S$-semimodule $M,$ we say that $I$ is $M$-injective ($M$-$i$-injective) [Gol1999, page 197], iff for every (normal) 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 $h\circ f=g.$

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

The following characterizations of semisimple rings are well known (cf., [Gri2007]):

Theorem 2.7**.**

([Gri2007, page 362, 402, 404])* Let $R$ be a ring. Then the following assertions are equivalent:*

1. (1)

Every left (right) $R$-module is $R$-injective; 2. (2)

Every left (right) $R$-module is injective; 3. (3)

Every left (right) $R$-module is projective; 4. (4)

Every short exact sequence of left (right) $R$-modules $0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0$ splits; 5. (5)

Every left (right) ideal of $R$ is a direct summand; 6. (6)

$R$* is left (right) semisimple.*

Lemma 2.8**.**

*([AN-1, Lemma 3.15.]) *If $M$ is a left $S$-semimodule such that every subtractive subsemimodule is a direct summand, then every left $S$-semimodule is $M$-$e$-projective.

2.9**.**

We say that a sequence of $S$-semimodules

[TABLE]

is

left splitting, iff there exists $f^{\prime}\in\mathrm{Hom}_{S}(B,A)$ such that $f^{\prime}\circ f=id_{A};$

right splitting, iff there exists $g^{\prime}\in\mathrm{Hom}_{S}(C,B)$ such that $g\circ g^{\prime}=id_{C}.$

splitting, iff it is left splitting and right splitting.

In 2009, a result to similar Theorem 2.7 was proved for subtractive semirings. We add a new characterization using $S$-$e$-projective semimodules.

Theorem 2.10**.**

If the semiring $S$ is left subtractive, then the following assertions are equivalent:

1. (1)

Every left $S$-semimodule is $S$-$e$-projective; 2. (2)

Every left $S$-semimodule is $S$-$k$-projective; 3. (3)

Every short exact sequence $0\rightarrow L\rightarrow S\rightarrow N\rightarrow 0$ of left $S$-semimodules is right splitting; 4. (4)

Every left ideal of $S$ is a direct summand; 5. (5)

$S$* is left ideal-semisimple.*

Proof**.**

The equivalences: (2) $\Leftrightarrow$ (4) $\Leftrightarrow$ (5) follow from [KNT2009, 4.4].

The equivalences (1) $\Leftrightarrow$ (3) is [AN-1, Proposition 3.14.].

(1) $\Rightarrow$ (2) follows from the fact that every $S$-$e$-projective left $S$-semimodule is $S$-$k$-projective.

(4) $\Rightarrow$ (1) This is Lemma 2.8 applied to $M=$ ${}_{S}S.\blacksquare$

For an arbitrary semiring, having every semimodule projective or injective or $e$-injective forces the ground semiring to be a semisimple ring. The following observation is a combination of [Ili2010, Theorem 3.1] and [AIKN2018, 5.3]:

Theorem 2.11**.**

The the following assertions are equivalent for any semiring $S:$

1. (1)

Every left (right) $S$-semimodule is projective; 2. (2)

Every left (right) $S$-semimodule is injective; 3. (3)

Every left (right) semimodule is $e$-injective; 4. (4)

$S$* is a left (right) semisimple ring.*

Our next goal is to find a relationship between the left ideal-semisimplicity of $S$ and having all left $S$-semimodules $S$-$e$-projective.

Definition 2.12**.**

Let $M$ be a left $S$-semimodule. A subsemimodule $N\leq_{S}M$ is a maximal summand of $M,$ iff $N\leq_{S}^{\oplus}M$ a direct summand of $M$ such that $N\neq M$ and for every direct summand $L\leq_{S}^{\oplus}M$ with $N\subseteq L\subseteq M,$ we have $N=L$ or $L=M.$ A direct summand $N\leq_{S}^{\oplus}M$ is called an irreducible summand, iff $\{0\}$ is a maximal direct summand of $N$.

Lemma 2.13**.**

([AN-4, 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.*

Lemma 2.14**.**

*([AN-4, 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]

Theorem 2.15**.**

If ${}_{S}S$ satisfies the ascending chain condition on direct summands, then $S=S_{1}\oplus\cdots\oplus S_{n}$, where $S_{i}$ is an irreducible summand for every $i\in\{1,\cdots,n\}$.

Proof**.**

By our assumptions and Lemma 2.13, $S$ satisfies also the descending chain condition on direct summands. If $S$ has no non-trivial direct summand, then [math] is the maximal summand of $S$, thus $S$ is an irreducible summand. If not, let $D_{0}$ be a non-trivial direct summand of $S$. Then

[TABLE]

is non-empty as $S\in\mathcal{D}_{1}$. Suppose that there exists $(D_{\lambda})_{\Lambda}$ a non-terminating descending chain in $\mathcal{D}_{1}$. Then there exists $\lambda_{i}\in\Lambda$, $i=0,1,2,\cdots$ such that $D_{\lambda_{0}}\supsetneqq D_{\lambda_{1}}\supsetneqq\cdots$, is a non-terminating strictly descending chain in $\mathcal{D}_{1}$, contradiction by the DCC on direct summands of ${}_{S}S$. Thus, the descending chain $(D_{\lambda})_{\Lambda}$ terminates and has a lower bound.

Since every descending chain in $\mathcal{D}_{1}$ has a lower bound, it follows by Zorn’s Lemma, that $\mathcal{D}_{1}$ has a minimal element, say $D_{1}$. Since there is no direct summand between $D_{0}$ and $D_{1}$, we see that $D_{0}$ is a maximal summand of $D_{1}$.

The set

[TABLE]

is non-empty as $0\in\mathcal{D}_{-1}$. Suppose that there exists $(D_{\lambda})_{\Lambda}$ a non-terminating ascending chain in $\mathcal{D}_{-1}$. Then there exist $\lambda_{i}\in\Lambda$, $i=0,1,\cdots$ such that $D_{\lambda_{0}}\subsetneqq D_{\lambda_{1}}\subsetneqq\cdots$, is a non-terminating ascending chain on $\mathcal{D}_{-1}$, contradiction by the ACC on direct summands of ${}_{S}S$. Thus the ascending chain $(D_{\lambda})_{\Lambda}$ terminates and has an upper bound.

Since every ascending chain on $\mathcal{D}_{-1}$ has an upper bound, it follows by Zorn’s Lemma, that $\mathcal{D}_{-1}$ has a maximal element say $D_{-1}$. Since there is no direct summand between $D_{-1}$ and $D_{0}$, we see that $D_{-1}$ is a maximal summand of $D_{0}$. We proved that every non-trivial direct summand is a maximal summand of a direct summand and has a maximal summand.

Now, let $D_{0}$ be a non-trivial direct summand of $S.$ Then there exists $D_{1},$ a direct summand of ${}_{S}S,$ such that $D_{0}$ is a maximal summand of $D_{1}$. If $D_{1}$ is non-trivial, then there exists $D_{2},$ a direct summand of $S,$ such that $D_{1}$ is a maximal summand of $D_{2}.$ Repeating this process over and over, we obtain an ascending chain

[TABLE]

of direct summands of ${}_{S}S,,$ which should terminate. Thus, there exists $n\in\mathbb{N}$ such that

[TABLE]

and $D_{i}$ is maximal summand of $D_{i+1}$ for $i=0,1,...,n-1$. Since $D_{0}$ is a non-trivial direct summand of $S$, $D_{0}$ has maximal summand $D_{-1}$. If $D_{-1}$ is non-trivial, then $D_{-1}$ has maximal summand $D_{-2}$. By repeating this process, we obtain a descending chain

[TABLE]

of direct summands of ${}_{S}S$, which should terminate. Thus, there exists $m\in\mathbb{N}$ such that

[TABLE]

and $D_{-i}$ is maximal summand of $D_{-i+1}$ for $i=1,2,...,m$. Hence

[TABLE]

is an ascending chain of direct summands of $S$ such that $D_{i}$ is a maximal summand of $D_{i+1}$ for $i=-m,-m+1,...,0,1,...,n-1$.

For $i=-m,-m+1,\cdots,0,1,\cdots,n-1$, write $S=D_{i}\oplus L_{i}$. Since $D_{i}\subsetneqq D_{i+1}$, we have

[TABLE]

with $D_{i+1}\cap L_{i}\neq 0$. Consider $K_{i+1}:=D_{i+1}\cap L_{i}.$ Then

[TABLE]

Suppose that there exists $i\in\{-m+1,-m+2,\cdots,n\}$ such that $K_{i}$ is a reducible summand. In this case, there exists a direct summand $K$ of $K_{i}$ such that $0\neq K\subsetneqq K_{i}$. Write $K_{i}:=K\oplus L$. Then

[TABLE]

thus $D_{i-1}\oplus K$ is a direct summand of $S$ such that

[TABLE]

contradiction to the maximality of $D_{i-1}$ as summand of $D_{i}$.$\blacksquare$

Remark 2.16**.**

If $S$ is a semiring with $S=\bigoplus\limits_{i\in I}N_{i},$ where $N_{i}$ is a non-zero left ideal of $S$ for every $i\in I$, then $I$ is finite. To see this, suppose that $I$ is infinite. Since $1\in S=\bigoplus\limits_{i\in I}N_{i}$ we have $1_{S}=\sum\limits_{j=1}^{k}n_{i_{j}}$ for some $k\in\mathbb{N},i_{j}\in I$ and $n_{i_{j}}\in N_{i_{j}}$. Let $i\in I\backslash\{i_{1},\cdots,i_{k}\}$ and $n_{i}\in N_{i}\backslash\{0\}.$ Then $n_{i}=n_{i}\cdot 1_{S}=n_{i}\cdot\sum\limits_{j=1}^{k}n_{i_{j}}=\sum\limits_{j=1}^{k}n_{i}n_{i_{j}}$, contradicting the uniqueness of the representation of $n_{i}$ in the direct sum.$\blacksquare$

Proposition 2.17**.**

Let $S$ be a semiring such that $S/I$ is $S$-$k$-projective for every subtractive ideal $I$ of $S.$

1. (1)

${}_{S}S$* satisfies the ACC on direct summands.* 2. (2)

$S=S_{1}\oplus S_{2}\oplus\cdots\oplus S_{n}$, where $S_{i}$ is an irreducible summand for every $i\in\{1,\cdots,n\}.$ If moreover, $S_{i}$ is ideal-simple (resp., congruence-simple) for every $i\in\{1,2,...,k\}$, then $S$ is ideal-semisimple (resp., congruence-semisimple).

Proof**.**

Assume that $S/I$ is $S$-$k$-projective for every subtractive ideal $I$ of $S.$

1. (1)

Suppose, without loss of generality, that there is a strictly ascending chain of direct summands of ${}_{S}S$:

[TABLE]

where, for each $i\in\mathbb{N}$ we have $S=N_{i}\oplus L_{i}$ for some left ideal $L_{i}\leq_{S}^{\oplus}S.$ Since, $N_{i}\subsetneqq N_{i+1},$we have $N_{i+1}\overset{\text{Lemma \ref{lemint}}}{{=}}N_{i}\oplus(N_{i+1}\cap L_{i})$ with $N_{i+1}\cap L_{i}\neq 0$ for each $i\in\mathbb{N}$. Setting $K_{1}:=N_{1}$ and $K_{i+1}:=N_{i+1}\cap L_{i}$ , for $i\geq 1,$ we have $N_{i}=K_{1}\oplus...\oplus K_{i}$ for every $i\geq 2$. Thus

[TABLE]

is a subtractive left ideal of $S$ as can be easily shown. So, we have an exact sequence of left $S$-semimodules

[TABLE]

Since $S/K$ is $S$-$k$-projective, there exists an $S$-linear map $\varphi:S/K\rightarrow S$ such that $\pi\circ\varphi=id_{S/K}$. For every $s\in S$, we have $\pi(\varphi([1]))=(\pi\circ\varphi)([1])=[1].$ Sine $\pi$ is $k$-normal, there exist $k,k^{\prime}\in K$ such that $1+k=\varphi([1])+k^{\prime}.$ Write $k=k_{1}+\cdots+k_{j}$ and $k^{\prime}=k_{1}^{\prime}+\cdots+k_{l}^{\prime},$ where $k_{i},k_{i}^{\prime}\in K_{i}$ for every $i,$ and let $m:=max\{j,l\}$. Then $k=k_{0}+k_{1}+...+k_{m}$ and $k^{\prime}=k_{1}^{\prime}+...+k_{m}^{\prime}$ for some $k_{i},k_{i}^{\prime}\in K_{i}$. Recall that for every $i\in\mathbb{N}$ we have

[TABLE]

For every $i\in\mathbb{N}$, let $\pi_{i}:S\rightarrow K_{i}$ be the canonical projection on $K_{i}$ and $e_{i}:=\pi_{i}(1).$ Then, $e_{i}=e_{i}1$ implies $\pi_{j}(e_{i})=\pi_{j}(e_{i}1)=e_{i}\pi_{j}(1)=e_{i}e_{j}$ and so $e_{i}e_{j}=0$ for every $i\neq j$ and $e_{i}e_{i}=e_{i}$. Since $k,k^{\prime}\in N_{m}$, we have $\pi_{m+1}(k)=0=\pi_{m+1}(k^{\prime})$. Thus

[TABLE]

Since $S=N_{m+1}\oplus L_{m+1}=K_{1}\oplus...\oplus K_{m}\oplus K_{m+1}\oplus L_{m+1}$, we have

[TABLE]

for some $l_{m+1}\in L_{m+1}$, whence

[TABLE]

Notice that

[TABLE]

It follows that $[1]=[e_{0}+e_{1}+...+e_{m}+l_{m+1}]$ while $\varphi([e_{0}+e_{1}+...+e_{m}+l_{m+1}])\neq\varphi([1])$, a contradiction. Hence, every ascending chain of direct summands of $S$ terminates, i.e. $S$ satisfies the ACC on direct summands. 2. (2)

By (1), the assumptions of Theorem 2.15 are satisfied, whence

[TABLE]

where $S_{i}$ is an irreducible summand for every $i\in\{1,\cdots,n\}$. If moreover, $S_{i}$ is ideal-simple (resp., congruence-simple) for every $i\in\{1,\cdots,n\}$, then $S$ is the direct sum of ideal-simple (resp. congruence-simple) left ideals, whence ideal-semisimple (resp. congruence-semisimple).$\blacksquare$

The following result is a combination of Lemma 2.8 and Proposition 2.17.

Corollary 2.18**.**

If $S$ is a semiring such that every subtractive left ideal is a direct summand, then $S=S_{1}\oplus\cdots\oplus S_{n}$, where $S_{i}$ is an irreducible summand for every $i\in\{1,\cdots,n\}.$ If moreover, $S_{i}$ is ideal-simple (resp., congruence-simple) for every $i\in\{1,\cdots,n\}$, then $S$ is ideal-semisimple (resp., congruence-semisimple).

Proposition 2.19**.**

For any semiring $S,$ each of the following conditions implies its successor:

1. (1)

Every subtractive ideal of $S$ is a direct summand. 2. (2)

Every $S$-semimodule is $S$-$e$-projective. 3. (3)

Every $S$-semimodule is $S$-$k$-projective. 4. (4)

$S/I$* is $S$-$k$-projective for every subtractive ideal $I$ of $S.$* 5. (5)

Every short exact sequence $0\longrightarrow I\longrightarrow S\longrightarrow N\longrightarrow 0$ in ${}_{S}\mathbf{SM}$ right splits. 6. (6)

${}_{S}S$* satisfies ACC on direct summands.* 7. (7)

${}_{S}S$* satisfies DCC on direct summands.* 8. (8)

$S=S_{1}\oplus\cdots\oplus S_{n}$, where every $S_{i}$ is an irreducible summand.

Proof**.**

(1) $\Rightarrow$ (2) This follows from Lemma 2.8 applied to $M=$ ${}_{S}S$. Let $M$ be an irreducible summand of ${}_{S}S,$ i.e. $\{{0\}}$ is the only maximal direct summand of ${}_{S}M$. By our assumption, $M\simeq M/{0}$ is ideal-simple.

(2) $\Rightarrow$ (3) $\Rightarrow$ (4) Follow directly from the definitions.

(4) $\Longleftrightarrow$ (5 ) Follows from [AN-1, Proposition 3.14] and Lemma 1.28.

(4) $\Rightarrow$ (6) Follows from Proposition 2.17.

(6) $\Leftrightarrow$ (7) Follows from Lemma 2.13.

(6) $\Rightarrow$ (8) Follows by Theorem 2.15.$\blacksquare$

Theorem 2.20**.**

([AN-4, 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.*

Proposition 2.21**.**

For any semiring $S,$ each of the following conditions implies its successor:

1. (1)

Every subtractive left ideal of $S$ is a direct summand. 2. (2)

Every left $S$-semimodule is $S$-$e$-injective. 3. (3)

Every $S$-semimodule is $S$-$i$-injective. 4. (4)

Every subtractive ideal of $S$ is $S$-$i$-injective. 5. (5)

Every short exact sequence $0\rightarrow L\rightarrow S\rightarrow N\rightarrow 0$ in ${}_{S}\mathbf{SM}$ is left splitting. 6. (6)

$S$* is $k$-Noetherian.* 7. (7)

$S$* satisfies the ACC on direct summands.* 8. (8)

$S$* satisfies the DCC on direct summands.* 9. (9)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand.

Proof**.**

(1) $\Rightarrow$ (2) Let $J$ be a left $S$-semimodule and let $f:M\rightarrow S$ a normal monomorphism, i.e. $M\leq S$ is a subtractive left ideal and $f$ is the canonical embedding. Let $g:M\rightarrow J$ be an $S$-linear map. By the assumption, $S=M\oplus N$ for some left ideal $N$ of $S.$ Let $\pi:S\rightarrow M$ be the projection on $M$ (i.e., $\pi\circ f=id_{M}$). Then $g\circ\pi:S\rightarrow J$ is an $S$-linear map satisfying $(g\circ\pi)\circ f=g.$

Let $h:M\rightarrow J$ be another $S$-linear map satisfying $h\circ f=g.$ Write $1_{S}=e_{M}+e_{N}$, where $e_{M}\in M$ are $e_{N}\in N$ are uniquely determined, and let $j_{0}:=h(1_{S}).$ For every $m\in M,$ we have $m=m1_{S}=m(e_{M}+e_{N})=me_{M}+me_{N}$, whence $me_{M}=m$ and $me_{N}=0$ as the sum $M+N$ is direct. Similarly, $ne_{M}=0$ and $ne_{N}=n$ for every $n\in N$. Define

[TABLE]

Then $(h_{1}\circ f)(m)=h_{1}(m)=me_{N}j_{0}=0$ for every $m\in M$. Moreover, we have

[TABLE]

Hence $J$ is $S$-$e$-injective.

The implications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) $\Rightarrow$ (5) \&\(6) $\Rightarrow$ (7) follow from the definitions.

(5) $\Rightarrow$ (6) Follows from Theorem 2.20.

(7) $\Longleftrightarrow$ (8) Follows from Lemma 2.13.

(7) $\Rightarrow$ (9) Follows from Theorem 2.15.$\blacksquare$

3 Commutative semisimple Semirings

The converse of Corollary 2.18 is satisfied when the semiring $S$ is commutative. To achieve this, we first prove the following technical result.

Lemma 3.1**.**

Let $S$ be a commutative ideal-semisimple (congruence-semisimple) semiring and write $S=S_{1}\oplus S_{2}\oplus...\oplus S_{k},$ where $S_{i}$ is an ideal-simple ideal of $S$ for every $i\in\{1,,\cdots,k\}$. Then every subtractive ideal $I$ of $S$ is a direct summand, and moreover $I=\bigoplus\limits_{a\in A}S_{a}$ for some $A\subseteq\{1,\cdots,k\}$.

Proof**.**

Let $I$ be a subtractive ideal of $S$ and

[TABLE]

Let $B:=\{1,\cdots,k\}\backslash A$ and write $S_{A}:=\bigoplus\limits_{a\in A}S_{a}$ and $S_{B}:=\bigoplus\limits_{b\in B}S_{b}$ . For every $a\in A$, the ideal $S_{a}$ is a (subtractive) ideal of $A,$ thus $I\cap S_{a}$ is a (subtractive) ideal. Since $0\neq I\cap S_{a}\subseteq S_{a}$ and $I\cap S_{a}$ is a (subtractive) left ideal, $I\cap S_{a}=S_{a}$. Thus $S_{A}\subseteq I$, and it follows that $I\overset{\text{Lemma \ref{lemint}}}{{=}}S_{A}\oplus(S_{B}\cap I)$.

Claim: $I\cap S_{B}=0$.

Let $1=e_{1}+...+e_{k}$ for some $e_{i}\in S_{i}$. For every $s_{i}\in S_{i}$,

[TABLE]

Since $s_{i}e_{j}\in S_{j}$ for every $j\in\{1,,\cdots,k\}$, it follows by the directness of the sum that $s_{i}e_{i}=s_{i}$ and $s_{i}e_{j}=0$ for every $i\neq j$. Therefore $e_{i}s_{i}=s_{i}$ and $e_{j}s_{i}=0$ for every $i\neq j$. Let $x\in I\cap S_{B}$, whence $x=\sum\limits_{b\in B}x_{b}$ where $x_{b}\in S_{b}$ for each $b\in B$. For every $\tilde{b}\in B$, we have $x_{\tilde{b}}=\sum\limits_{b\in B}e_{\tilde{b}}x_{b}=e_{\tilde{b}}x\in I$ as $I$ is an ideal. Thus $x_{b}=0$ for every $b\in B$ and $x=0$.$\blacksquare$

The following technical conditions shall be needed in the sequel.

3.2**.**

Let $N$ be a left $S$-semimodule. Consider the conditions:

$\mathbf{C1:\ }$Every subtractive $S$-subsemimodule $M\leq_{S}N$ is a direct summand.

$\mathbf{C2:\ }$For every subtractive $S$-subsemimodule $M\leq_{S}N$ and every maximal subtractive $S$-subsemimodule $L\leq_{S}M$, the left $S$-semimodule $M/L$ is left ideal-simple.

$\mathbf{C2}^{\prime}\mathbf{:}$ For every subtractive subsemimodule $M\leq_{S}N$ and every maximal subtractive $S$-subsemimodule $L\leq_{S}M$, the left $S$-semimodule $M/L$ is congruence-simple.

Remark 3.3**.**

The conditions $\mathbf{C1}$ and $\mathbf{C2}$ (and $\mathbf{C2}^{\prime}$) are independent:

1. (1)

$B(3,2)$ satisfies $\mathbf{C1}$ but neither $\mathbf{C2}$ nor $\mathbf{C2}^{\prime}.$ 2. (2)

$B(3,1)$ satisfies $\mathbf{C2}$ but not $\mathbf{C1}.$ 3. (3)

$\mathbb{B}^{\mathbb{N}}$ satisfies $\mathbf{C2^{\prime}}$ but not $\mathbf{C1}.$ 4. (4)

$\mathbb{R}^{+}$ satisfies $\mathbf{C2}$ but not $\mathbf{C2^{\prime}.}$ By Example 1.16, $\mathbb{R}^{+}$ is ideal-simple but not congruence-simple. Since $\mathbb{R}^{+}$ is ideal-simple, it has no proper non-trivial ideals, $\{0\}$ is the maximal subtractive subsemimodule of $\mathbb{R}^{+},$ and $\mathbb{R}^{+}/\{0\}\simeq\mathbb{R}^{+}$ is ideal-simple. Hence $\mathbb{R}^{+}$ satisfies $\mathbf{C2}$. However, $\mathbb{R}^{+}/\{0\}\simeq\mathbb{R}^{+}$ is not congruence-simple, thus $\mathbb{R}^{+}$ does not satisfy $\mathbf{C2^{\prime}.}$ 5. (5)

Let $(M,+,0)$ be a finite lattice which is not distributive. $\mathbf{E}_{M},$ the endomorphism semiring of $M,$ satisfies $\mathbf{C2^{\prime}}$ but not $\mathbf{C2.}$ By Example 1.15, $\mathbf{E}_{M}$ is left congruence-simple but not left ideal-simple. Since $\mathbf{E}_{M}$ is left congruence-simple, it has no non-trivial subtractive left ideals, $\{0\}$ is the maximal subtractive ideal of $\mathbf{E}_{M}$ and $\mathbf{E}_{M}/\{0\}=\mathbf{E}_{M}$ is left congruence-simple. Hence, $\mathbf{E}_{M}$ satisfies $\mathbf{C2^{\prime}}$. However, $\mathbf{E}_{M}/\{0\}=\mathbf{E}_{M}$ is not ideal-simple, thus $\mathbf{E}_{M}$ does not satisfy $\mathbf{C2}$.

The following result extends the characterizations of ideal-semisimple semirings in Theorem 2.10 to commutative not necessarily subtractive semirings:

Theorem 3.4**.**

The following assertions are equivalent for a commutative semiring $S$:

1. (1)

Every subtractive ideal of $S$ is a direct summand and $S$ satisfies $\mathbf{C2};$ 2. (2)

Every $S$-semimodule is $S$-$e$-projective and $S$ satisfies $\mathbf{C2};$ 3. (3)

Every $S$-semimodule is $S$-$k$-projective and $S$ satisfies $\mathbf{C2};$ 4. (4)

$S/I$* is $S$-$k$-projective for every subtractive ideal $I$ of $S,$ and $S$ satisfies $\mathbf{C2};$* 5. (5)

Every short exact sequence $0\longrightarrow I\longrightarrow S\longrightarrow N\longrightarrow 0$ in ${}_{S}\mathbf{SM}$ right splits and $S$ satisfies $\mathbf{C2};$ 6. (6)

${}_{S}S$* satisfies the ACC on direct summands and $\mathbf{C2};$* 7. (7)

${}_{S}S$* satisfies the DCC on direct summands and $\mathbf{C2};$* 8. (8)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand, and $S$ satisfies $\mathbf{C2};$ 9. (9)

$S$* is ideal-semisimple.*

Proof**.**

By Proposition 2.19, we only need to prove (8) $\Rightarrow$ (9) and (9) $\Rightarrow$ (1).

(8) $\Rightarrow$ (9) Notice that assuming $\mathbf{C2}$ guarantees that $S_{i}$ is ideal-simple for $i=1,\cdots,n.$ Whence, $S$ is ideal-semisimple.

(9) $\Rightarrow$ (1) Assume that $S$ is ideal-semisimple and write $S=S_{1}\oplus\cdots\oplus S_{k}$ for some $k\in\mathbb{N}$ with $S_{i}$ an ideal-simple ideal for $i=1,\cdots,k$. Let $I$ be a subtractive ideal of $S$. Since $S$ is commutative, it follows by Lemma 3.1 that $I=\bigoplus\limits_{a\in A}S_{a}$ for some $A\subseteq\{1,\cdots,k\}$, whence $S\overset{\text{Lemma \ref{lemint}}}{{=}}I\oplus\bigoplus\limits_{b\notin A}S_{b}$. Hence, $I$ is a direct summand of $S$.

Claim: ${}_{S}S$ satisfies ${\mathbf{C2}}$.

Let $M$ be a subtractive ideal of $S$ and $L$ a maximal subtractive subideal of $M$. Then $M=S_{A}=\bigoplus\limits_{a\in A}S_{a}$ and $L=S_{C}=\bigoplus\limits_{c\in C}S_{c}$ for some $C\subsetneqq A\subseteq\{1,\cdots,k\}$. Notice that $C\subseteq A$ since $L\subseteq M$. Moreover, $|A\backslash C|=1$ since $|A\backslash C|=0$ implies $L=M$ and $\left|A\backslash C\right|\geq 2$ implies $L\subsetneqq S_{C\cup\{y\}}\subsetneqq M$ for some $y\in A\backslash C$ with $S_{C\cup\{y\}}$ a subtractive ideal of $S$, contradiction to the maximality of $L$. Write $A\backslash C=\{x\}$ and $B=\{1,2,...,k\}\backslash A$. Then $S=S_{A}\oplus S_{B}=S_{C}\oplus S_{x}\oplus S_{B}$ where $S_{B}=\bigoplus\limits_{b\in B}S_{b}$.

Let $I$ be an ideal of $S$ such that $L\subsetneqq I\subseteq M.$ Then there exists $i\in I\backslash N$. Since $i\in M$, $i=t_{C}+t_{x}$ for some $t_{C}\in S_{C},t_{x}\in S_{x}$. Notice that $t_{x}\neq 0$; otherwise, $i=t_{C}\in N$. Moreover, $0\neq t_{x}=e_{x}t_{x}=e_{x}(t_{C}+t_{x})=e_{x}i\in I$. Thus $I\cap S_{x}\neq 0$, whence $I\cap S_{x}=S_{x}$ as $S_{x}$ is ideal-simple. Since $S_{C}\subseteq I$ and $S_{x}\subseteq I$, we have $M=S_{C}+S_{x}\subseteq I$.$\blacksquare$

The following result is the “congruence-semisimple” version of Theorem 3.4.

Theorem 3.5**.**

The following assertions are equivalent for a commutative semiring $S$:

1. (1)

Every subtractive ideal of $S$ is a direct summand and $S$ satisfies $\mathbf{C2}^{\prime};$ 2. (2)

Every $S$-semimodule is $S$-$e$-projective and $S$ satisfies $\mathbf{C2}^{\prime};$ 3. (3)

Every $S$-semimodule is $S$-$k$-projective and $S$ satisfies $\mathbf{C2}^{\prime};$ 4. (4)

$S/I$* is $S$-$k$-projective for every subtractive ideal $I$ of $S$ and $S$ satisfies $\mathbf{C2}^{\prime};$* 5. (5)

Every short exact sequence $0\longrightarrow I\longrightarrow S\longrightarrow N\longrightarrow 0$ right splits and $S$ satisfies $\mathbf{C2}^{\prime};$ 6. (6)

${}_{S}S$* satisfies the ACC on direct summands and $S$ satisfies $\mathbf{C2}^{\prime};$* 7. (7)

${}_{S}S$* satisfies the DCC on direct summands and $S$ satisfies $\mathbf{C2}^{\prime};$* 8. (8)

$S=S_{1}\oplus\cdots\oplus S_{n}$, where every $S_{i}$ is an irreducible summand, and $S$ satisfies $\mathbf{C2}^{\prime};$ 9. (9)

$S$* is congruence-semisimple.*

Proof**.**

We only need to prove (9) $\Rightarrow$ (1); the proof of the other implications are similar to the proof of the corresponding ones in Theorem 3.4.

Assume that $S$ is congruence-semisimple. With the help of Lemma 3.1, it can be shown, as in the proof of Theorem 3.4, that every subtractive ideal of $S$ is a direct summand.

Claim: $S$ satisfies $\mathbf{C2}^{\prime}\mathbf{.}$

Let $M,$ $L$ be subtractive ideals of $S$ with $L$ a maximal subtractive $S$-subsemimodule of $M$. Then similarly to the proof of Theorem 3.4, we have $M=S_{A}:=\bigoplus\limits_{a\in A}S_{a}$, $S=S_{A}\oplus S_{B}$ and $L=S_{C}:=\bigoplus\limits_{c\in C}S_{c}$ where $C\cup\{x\}=A$.

Let $\rho$ be a congruence relation on $S$ such that $\equiv_{L}\subsetneqq\rho\subseteq\equiv_{M}.$ Consider the congruence relation $\rho^{\prime}$ on $S_{x}$ defined by

[TABLE]

**Step I: ** $\rho^{\prime}=S_{x}^{2}$.

Since $\equiv_{N}\neq\rho$, there exist $s,s^{\prime}\in S$ such that $s\not\equiv_{L}s^{\prime}$ and $s\rho s^{\prime}$. Write $s=s_{C}+s_{x}+s_{B}$ and $s^{\prime}=s_{C}^{\prime}+s_{x}^{\prime}+s_{B}^{\prime}$ for some $s_{C},s_{C}^{\prime}\in S_{C},s_{x},s_{x}^{\prime}\in S_{x},s_{B},s_{B}^{\prime}\in S_{B}$. Since $s\equiv_{M}s^{\prime}$, there exists $m,m^{\prime}\in M=S_{A}$ such that $m+s=m^{\prime}+s^{\prime}$, that is $(m+s_{C}+s_{x})+s_{B}=(m^{\prime}+s_{C}^{\prime}+s_{x}^{\prime})+s_{B}^{\prime}$ whence $s_{B}=s_{B}^{\prime}$ as the sum $S_{A}+S_{B}$ is direct. Notice that $s_{x}\neq s_{x}^{\prime}$; otherwise, $s_{C}^{\prime}+s=s_{C}+s^{\prime}$ where $s_{C},s_{C}^{\prime}\in S_{C}=L$, a contradiction (with $\equiv_{L}\neq\rho$). Therefore, $s_{x}\rho^{\prime}s_{x}^{\prime}$ and $s_{x}\neq s_{x}^{\prime}$, whence $\rho^{\prime}=S_{x}^{2}$ as $S_{x}$ is congruence-simple.

**Step II: ** $\rho=\equiv_{M}$.

Let $s,s^{\prime}\in S$ be such that $s\equiv_{M}s^{\prime}$ and write $s=s_{C}+s_{x}+s_{B}$, $s^{\prime}=s_{C}^{\prime}+s_{x}^{\prime}+s_{B}^{\prime}$ for some $s_{C},s_{C}^{\prime}\in S_{C},s_{x},s_{x}^{\prime}\in S_{x},s_{B},s_{B}^{\prime}\in S_{B}$. Then $s_{B}=s_{B}^{\prime}$. Since $\rho^{\prime}=S_{x}^{2}$, we have $s_{x}\rho^{\prime}s_{x}^{\prime}$, whence $(t_{C}+s_{x}+t_{B})\rho(t_{C}^{\prime}+s_{x}^{\prime}+t_{B}^{\prime})$ for some $t_{C},t_{C}^{\prime}\in S_{C},t_{B},t_{B}^{\prime}\in S_{B}$. Thus $e_{x}(t_{C}+s_{x}+t_{B})\rho e_{x}(t_{C}^{\prime}+s_{x}^{\prime}+t_{B}^{\prime})$, that is $s_{x}\rho s_{x}^{\prime}$. Since $s_{C}\equiv_{L}s_{C}^{\prime},$ we have $s_{x}\rho s_{x}^{\prime}$, and $s_{B}=s_{B}^{\prime}$, $(s_{C}+s_{x}+s_{B})\rho(s_{C}^{\prime}+s_{x}^{\prime}+s_{B}^{\prime})$, that is $s\rho s^{\prime}$. We conclude that $\rho=\equiv_{M}$.$\blacksquare$

Definition 3.6**.**

The semiring $S$ is

left (right) $k$-Noetherian, iff every ascending chain of subtractive left (right) ideals of $S$ terminates;

left (right) $k$-Artinian, iff every descending chain of subtractive left (right) ideals of $S$ terminates.

Theorem 3.7**.**

([AN-4, Theorem 2.13])* If every subtractive left ideal of $S$ is a direct summand, then $S$ is left $k$-Artinian and left $k$-Noetherian.*

The following result is a combination of Theorems 3.4, 3.5 and 3.7:

Corollary 3.8**.**

If $S$ is a commutative ideal-semisimple (congruence-semisimple) semiring, then $S$ is $k$-Artinian and $k$-Noetherian.

The following examples show that $\mathbf{C2}$ (resp., $\mathbf{C2}^{\prime}$) cannot be dropped from the assumptions of Theorem 3.4 (resp., Theorem 3.5).

Example 3.9*.*

Consider the commutative semiring $B(p+1,p),$ where $p$ is an odd prime number.

1. (1)

Every subtractive ideal is a direct summand. 2. (2)

Every $B(p+1,p)$-semimodule is $B(p+1,p)$-$e$-projective. 3. (3)

$B(p+1,p)$ is not left ideal-semisimple. 4. (4)

$B(p+1,p)$ is not congruence-semisimple.

Proof**.**

Notice that the only ideals of $B(p+1,p)$ are $\{0\},$ $B(p+1,p)$, and $I=\{0,p\}.$

1. (1)

The only subtractive ideals of $B(p+1,p)$ are $\{0\}$ and $B(p+1,p),$ each of which is a direct summand of $B(p+1,p).$ 2. (2)

Since (1) is valid, it follows by Lemma 2.8, that all $B(p+1,p)$-ideals are $B(p+1,p)$-$e$-projective. 3. (3)

$B(p+1,p)$ is an irreducible summand which is not ideal-simple (it contains the ideal $I$). So, $B(p+1,p)$ is not ideal-semisimple. Notice that $B(p+1,p)$ does not satisfy $\mathbf{C2}$. 4. (4)

$B(p+1,p)$ is not an irreducible summand which is not congruence-simple ($\rho=\{(i,j)|$ $i,j\neq 0\}$ is a non trivial congruence relation on $B(p+1,p)$). Notice that $B(p+1,p)$ does not satisfy $\mathbf{C2}^{\prime}$.$\blacksquare$

Example 3.10*.*

Consider the semiring $S:=B(3,1).$

1. (1)

$I:=\{0,2\}$ is a subtractive ideal of $B(3,1)$, which is not a direct summand of $B(3,1)$; 2. (2)

$B(3,1)$ is not ideal-semisimple; 3. (3)

$B(3,1)$ is not congruence-semisimple.

Proof**.**

Notice that the only ideals of $S$ are $0,$ $I$ and $S,$ which are subtractive. Moreover, $I$ is the maximal subtractive subsemimodule of $S$ and is clearly not a direct summand of $S.$ Moreover, $\{0_{S}\}$ is the maximal subtractive ideal of $I.$ Notice that $I/0\cong\mathbb{B}\cong S/I$ as $S$-semimodules, whence $I/0$ and $S/I$ are ideal-simple. Thus $S$ is an irreducible summand that is neither ideal-simple ($I$ is a non trivial left ideal of $S$ ) nor congruence-simple ($\equiv_{I}$ is a non trivial congruence relation of $S$).

Theorem 3.11**.**

Let $S$ be a commutative semiring which satisfies $\mathbf{C2}.$ The following assertions are equivalent:

1. (1)

Every subtractive ideal of $S$ is a direct summand; 2. (2)

Every $S$-semimodule is $S$-$e$-injective; 3. (3)

Every $S$-semimodule is $S$-$i$-injective; 4. (4)

Every subtractive ideal of $S$ is $S$-$i$-injective; 5. (5)

Every short exact sequence $0\rightarrow L\rightarrow S\rightarrow N\rightarrow 0$ in ${}_{S}\mathbf{SM}$ is left splitting; 6. (6)

$S$* is $k$-Noetherian;* 7. (7)

$S$* satisfies the ACC on direct summands;* 8. (8)

$S$* satisfies the DCC on direct summands;* 9. (9)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand; 10. (10)

$S$* is ideal-semisimple.*

Proof**.**

This is a consequence of Proposition 2.21 and the proof of Theorem 3.4.$\blacksquare$

Combining Theorems 3.4 and 3.11, we obtain the following characterization of commutative ideal-semisimple semirings:

Theorem 3.12**.**

The following assertions are equivalent for a commutative semiring $S$ which satisfies $\mathbf{C2}$:

1. (1)

Every subtractive ideal of $S$ is a direct summand; 2. (2)

Every $S$-semimodule is $S$-$e$-projective ($S$-$k$-projective); 3. (3)

Every $S$-semimodule is $S$-$e$-injective ($S$-$i$-injective); 4. (4)

For every subtractive ideal $I$ of $S$ we have: $S/I$ is $S$-$k$-projective ($I$* is $S$-$i$-injective);* 5. (5)

Every short exact sequence $0\longrightarrow I\longrightarrow S\longrightarrow N\longrightarrow 0$ in ${}_{S}\mathbf{SM}$ right splits (left splits); 6. (6)

$S$* is $k$-Noetherian;* 7. (7)

${}_{S}S$* satisfies ACC on the direct summands;* 8. (8)

${}_{S}S$* satisfies DCC on the direct summands;* 9. (9)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand; 10. (10)

$S$* is ideal-semisimple.*

The following result is the congruence-semisimple version of Theorem 3.11.

Theorem 3.13**.**

Let $S$ be a commutative semiring which satisfies $\mathbf{C2}^{\prime}.$ The following assertions are equivalent:

1. (1)

Every subtractive ideal of $S$ is a direct summand; 2. (2)

Every $S$-semimodule is $S$-$e$-injective; 3. (3)

Every $S$-semimodule is $S$-$i$-injective; 4. (4)

Every subtractive ideal of $S$ is $S$-$i$-injective; 5. (5)

Every short exact sequence of $S$-semimodules $0\rightarrow L\rightarrow S\rightarrow N\rightarrow 0$ is left splitting; 6. (6)

$S$* is $k$-Noetherian;* 7. (7)

$S$* satisfies ACC on direct summands;* 8. (8)

$S$* satisfies DCC on direct summands;* 9. (9)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand; 10. (10)

$S$* is congruence-semisimple.*

Combining Theorems 3.5 and 3.13, we obtain the following characterization of commutative congruence-semisimple semirings:

Theorem 3.14**.**

The following assertions are equivalent for a commutative semiring $S$ which satisfies $\mathbf{C2}^{\prime}$:

1. (1)

Every subtractive ideal of $S$ is a direct summand; 2. (2)

Every $S$-semimodule is $S$-$e$-projective ($S$-$k$-projective); 3. (3)

Every $S$-semimodule is $S$-$e$-injective ($S$-$i$-injective); 4. (4)

For every subtractive ideal $I$ of $S$ we have: $S/I$ is $S$-$k$-projective ($I$* is $S$-$i$-injective);* 5. (5)

Every short exact sequence $0\longrightarrow I\longrightarrow S\longrightarrow N\longrightarrow 0$ in ${}_{S}\mathbf{SM}$ right splits (left splits); 6. (6)

$S$* is $k$-Noetherian;* 7. (7)

${}_{S}S$* satisfies the ACC on the direct summands;* 8. (8)

${}_{S}S$* satisfies the DCC on the direct summands;* 9. (9)

$S=S_{1}\oplus S_{2}\oplus S_{3}\oplus...\oplus S_{n}$, where every $S_{i}$ is an irreducible summand; 10. (10)

$S$* is congruence-semisimple.*

The following example shows that the assumption that $S$ is a *commutative *semiring cannot be dropped from Theorems 3.4, 3.11, whence not from our main result Theorems 3.12, 3.14:

Example 3.15*.*

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

1. (1)

$S$ is a left ideal-semisimple semiring. 2. (2)

$N_{1}$ is a subtractive left ideal of $S$ which is not a direct summand. 3. (3)

$S/N_{1}$ is not an $S$-$k$-projective $S$-semimodule (whence not $S$-$e$-projective). 4. (4)

$N_{1}$ is not $S$-$e$-injective.

Proof**.**

1. (1)

The semiring $M_{2}(\mathbb{R}^{+})$ is left ideal-simple since $\mathbb{R}^{+}$ is a semifield ([HW1998, Theorem 7.8]). 2. (2)

Let $K$ be a left ideal of $S$ such that $S=N_{1}+K$. Then $1_{S}=i+k$ for some $i\in N_{1}$ and $k\in K$, that is

[TABLE]

Then $p+a=1=s+b$ and $q+a=0=r+b$, whence $a=q=r=b=0$ as $\mathbb{R}^{+}$ is zerosumfree. Therefore, $i=0$ and $k=1_{S}$, which implies $K=S$ and $0\neq N_{1}=N_{1}\cap K.$ Thus, the sum $N_{1}+K$ is not direct. Consequently, $N_{1}$ is a subtractive left ideal of $S$ which is not a direct summand. 3. (3)

Let $\pi:S\rightarrow S/N_{1}$ be the canonical map and $id_{S/N_{1}}$ be the identity map of $S/N_{1}$. Notice that $\pi$ is a normal epimorphism. Consider

[TABLE]

Suppose that there exists an $S$-linear map $g:S/N_{1}\rightarrow S$ such that $\pi\circ g=id_{S/N_{1}}$. Then $g(\overline{e_{1}})\in\pi^{-1}(\overline{e_{1}})$ and $g(\overline{e_{2}})\in\pi^{-1}(\overline{e_{2}})$. Write g(\overline{e_{1}}):=\left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right] for some $p,q,r,s\in\mathbb{R}^{+}$. Then \left[{\begin{array}[]{cc}p+k&q+k\\ r+l&s+l\end{array}}\right]=\left[{\begin{array}[]{cc}m+1&m\\ n&n\end{array}}\right] for some $k,l,m,n\in\mathbb{R}^{+}$, whence $r=s$ and $p=q+1$ as $\mathbb{R}^{+}$ is cancellative. By relabeling, we have g(\overline{e_{1}})=\left[{\begin{array}[]{cc}a+1&a\\ b&b\end{array}}\right] for some $a,b\in\mathbb{R}^{+}$. Similarly, g(\overline{e_{2}})=\left[{\begin{array}[]{cc}c&c\\ d&d+1\end{array}}\right] for some $c,d\in\mathbb{R}^{+}$.

Let x:=\left[{\begin{array}[]{cc}p&q\\ r&s\end{array}}\right]\in S. Then x=\left[{\begin{array}[]{cc}p&0\\ r&0\end{array}}\right]e_{1}+\left[{\begin{array}[]{cc}0&q\\ 0&s\end{array}}\right]e_{2}, whence

[TABLE]

But x=\left[{\begin{array}[]{cc}p&1\\ r&0\end{array}}\right]e_{1}+\left[{\begin{array}[]{cc}0&q\\ 1&s\end{array}}\right]e_{2}, whence

[TABLE]

whence $b=0=c$ as $\mathbb{R}^{+}$ is cancellative. Thus g(\overline{e_{1}})=\left[{\begin{array}[]{cc}a+1&a\\ 0&0\end{array}}\right] for some $a,b\in\mathbb{R}^{+}$ and g(\overline{e_{2}})=\left[{\begin{array}[]{cc}0&0\\ d&d+1\end{array}}\right].

Let y=\left[{\begin{array}[]{cc}2&1\\ 0&0\end{array}}\right]. Notice that $\overline{e_{1}}=\overline{y}$, whence

[TABLE]

and so $a=2a+d+1$. Since $\mathbb{R}^{+}$ is cancellative, $a+d+1=0$, that is $1$ has an additive inverse, a contradiction. Hence, there is no such $S$-linear map $g$ with $\pi\circ g=id_{S/I};$ i.e., $S/I$ is not $S$-$k$-projective. Since $S/I$ is not $S$-$k$-projective, $S/I$ is not $S$-$e$-projective. 4. (4)

This was shown in [AN-2, Example 2.19.].$\blacksquare$