On the idempotent semirings such that $\mathcal{D}^\bullet$ is the least distributive lattice congruence

TL;DR

This paper characterizes the least distributive lattice congruence on idempotent semirings and describes the structure of specific varieties where this congruence coincides with known relations.

Contribution

It provides a general description of the least distributive lattice congruence on idempotent semirings and characterizes the varieties where this congruence matches specific known congruences.

Findings

The least distributive lattice congruence $ heta$ on an idempotent semiring is characterized.

Varieties $D^ullet, L^ullet, R^ullet$ are described where $ heta$ equals $ ext{D}^ullet, ext{L}^ullet, ext{R}^ullet$.

Semirings in $D^ullet$ are spined products of semirings in $L^ullet$ and $R^ullet$ with respect to a distributive lattice.

Abstract

Here we describe the least distributive lattice congruence $\eta$ on an idempotent semiring in general and characterize the varieties $D^\bullet, L^\bullet$ and $R^\bullet$ of all idempotent semirings such that $\eta=\mathcal{D}^\bullet, \mathcal{L}^\bullet$ and $\mathcal{R}^\bullet$, respectively. If $S \in D^\bullet [L^\bullet, R^\bullet]$, then the multiplicative reduct $(S, \cdot)$ is a [left, right] normal band. Every semiring $S \in D^\bullet$ is a spined product of a semiring in $L^\bullet$ and a semiring in $R^\bullet$ with respect to a distributive lattice.

Tables

Notation	Defining identity within $𝐈$	Notation	Defining identity within $𝐈$
${𝐑}^{+}$	$x+y+x\approx x$,	${𝐑}^{\bullet }$	$xyx\approx x$,
${\mathrm{𝐋𝐙}}^{+}$	$x+y\approx x$,	${\mathrm{𝐑𝐙}}^{+}$	$x+y\approx y$,
${\mathrm{𝐋𝐙}}^{\bullet }$	$xy\approx x$,	${\mathrm{𝐑𝐙}}^{\bullet }$	$xy\approx y$,
${\mathrm{𝐋𝐍𝐁}}^{\bullet }$	$xyz\approx xzy$,	${\mathrm{𝐑𝐍𝐁}}^{\bullet }$	$xyz\approx yxz$,
$\mathrm{𝐋𝐐𝐁𝐈}$	$x+xy+x\approx x$,	$\mathrm{𝐑𝐐𝐁𝐈}$	$x+yx+x\approx x$,
$\mathrm{𝐋𝐍}$	$x+xyx\approx x$,	$\mathrm{𝐑𝐍}$	$xyx+x\approx x$,
$𝐍$	$x+xyx+x\approx x$,	$𝒮l^{+}$	$x+y\approx y+x$.

Equations

$$x(y+z)=xy+xz\text{and}(x+y)z=xz+yz.$$

$$x+x\approx x\text{and}x\cdot x\approx x.$$

$$a{\mathcal{D}}^{\bullet }b\iff aba=a,bab=b,$$

$$a{\mathcal{L}}^{\bullet }b\iff ba=b,ab=a$$

$$a{\mathcal{R}}^{\bullet }b\iff ab=b,ba=a.$$

$$x+xy+x$$

$$\text{and}x+yx+x$$

$$x+yxy\approx x;$$

$$a\sigma b$$

$$aba=aba+a+aba\text{and}bab=bab+b+bab.$$

$$a{\sigma }^{\ast }b$$

$$axbxa=axbxa+a+axbxa\text{and}bxaxb=bxaxb+b+bxaxb.$$

$$xz+xyz+xz\approx xz.$$

$$aba$$

$$=aba+bxa+aba$$

$$=aba+(ab+axb+ab)xa+aba$$

$$=aba+abxa+axbxa+abxa+aba$$

$$=aba+abxa+axbxa+a+axbxa+abxa+aba$$

$$=aba+abxa+axbxa+a+(aba+abxa+$$

$$axbxa+a+axbxa+abxa+aba)$$

$$=aba+abxa+axbxa+a+aba$$

$$=aba+a+aba$$

$$a{\le }_{+}^{l}b\iff b=a+ba{\le }_{\cdot }^{l}b\iff a=ba$$

$$a{\le }_{+}^{r}b\iff b=b+aa{\le }_{\cdot }^{r}b\iff a=ab$$

$${\le }_{+}={\le }_{+}^l\cap {\le }_{+}^r\text{and}{\le }_{\cdot }={\le }_{\cdot }^l\cap {\le }_{\cdot }^r$$

$$x\approx xyx+x+xyx.$$

$$a=a(a+ab)a\Rightarrow$$

$$\Rightarrow$$

$$aba=aba+a+aba\text{and}bab=bab+b+bab.$$

$$aba+ab+aba=aba$$

$$\Rightarrow$$

$$x\approx xy+x+xy;$$

$$x\approx x(y+x+y).$$

$$a+b=(a+b)(b+a)=ab+a+b+ba$$

$$b+a=(b+a)(a+b)=ba+b+a+ab$$

$$a=a(a+b)+a(b+a)$$

$$\Rightarrow$$

Full text

a) Department of Pure Mathematics, University of Calcutta, Kolkata, India

b) Department of Mathematics, Visva-Bharati, Santiniketan-731235, India.

c) Department of Mathematics, Kurseong college, Kurseong-734203, India.

E-mail addresses: senmk6@yahoo.coma, anjankbhuniya@gmail.comb and rajib.d6@gmail.comc.

Keywords : idempotent semiring; least distributive lattice congruence; normal band, spined product, Malcev’s product.

AMS Mathematics Subject Classification: 16Y60.

1 Introduction

A semiring $(S,+,\cdot)$ is an algebra with two binary operations + and $\cdot$ such that both the additive reduct $(S,+)$ and the multiplicative reduct $(S,\cdot)$ are semigroups and such that the following distributive laws hold:

[TABLE]

If moreover, both the reducts $(S,+)$ and $(S,\cdot)$ are bands, then $S$ is called an idempotent semiring. Thus the class of all idempotent semirings is an equational class satisfying two additional identities:

[TABLE]

The variety of all idempotent semirings will be denoted by $\mathbf{I}$.

We now fix the notation for some varieties of idempotent semirings and give their determining identity within $\mathbf{I}$:

For an idempotent semiring $(S,+,\cdot)$ the Green’s relations $\mathcal{L},\mathcal{R}$ and $\mathcal{D}$ on the additive [multiplicative] reduct $(S,+)[(S,\cdot)]$ will be denoted by $\mathcal{L}^{+},\mathcal{R}^{+}$ and $\mathcal{D}^{+}[\mathcal{L}^{\bullet},\mathcal{R}^{\bullet},\mathcal{D}^{\bullet}]$. Since the multiplicative reduct $(S,\cdot)$ of an idempotent semiring $S$ is a band, $\mathcal{D}^{\bullet},\mathcal{L}^{\bullet}$ and $\mathcal{R}^{\bullet}$ are given by: for $a,b\in S$,

[TABLE]

The relations $\mathcal{D}^{+},\mathcal{L}^{+}$ and $\mathcal{R}^{+}$ are given dually.

The variety $\mathbf{I}$ of all idempotent semirings contains the variety $\mathbf{D}$ of all distributive lattices; and hence there is the least distributive lattice congruence on each idempotent semiring $S$. Throughout this article, we denote the least distributive lattice congruence on $S$ by $\eta_{S}$ or precisely by $\eta$. Then $(S/\eta,+)$ is a semilattice. Since $(S,+)$ is a band, $\mathcal{D}^{+}$ is the least congruence on the idempotent semiring $S$ such that $(S/\mathcal{D}^{+},+)$ is a semilattice. Hence $\mathcal{D}^{+}\subseteq\eta$. In [13], Sen, Guo and Shum characterized in depth the idempotent semirings $S$ such that $\eta=\mathcal{D}^{+}$.

Lemma 1.1**.**

[13]** Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$\mathcal{D}^{+}$* is the least distributive lattice congruence on $S$;* 2. 2.

$S$* satisfies the identities:*

[TABLE] 3. 3.

$S\in\mathbf{R}^{+}\circ\mathbf{D}$.

An idempotent semiring $S$ is said to be a band semiring if $\mathcal{D}^{+}$ is the least distributive lattice congruence on $S$, equivalently if it satisfies the identities (1.1) and (1.2). We will denote the variety of all band semirings by $\mathbf{BI}$. We define an idempotent semiring to be a left [right] quasi-band semiring if it satisfies the identity (1.1) [(1.2)]. Thus $\mathbf{BI}=\mathbf{LQBI}\cap\mathbf{RQBI}$. Additive reduct $(S,+)$ of a band semiring $S$ is a regular band [14].

The idempotent semirings for which $\eta=\mathcal{D}^{\bullet}$ have been characterized by Pastijn and Zhao [9]. There are several articles characterizing the variety $\mathbf{I}$ of idempotent semirings by Green’s relations [7],  [9], [13] - [16].

We extend the problem to the idempotent semirings $S$ such that the least distributive lattice congruence $\eta$ on $S$ is either of the Green’s relations $\mathcal{L}^{+}$, $\mathcal{R}^{+}$, $\mathcal{L}^{\bullet}$ and $\mathcal{R}^{\bullet}$. For this, we first describe explicitly the least distributive lattice congruence $\eta$ on an idempotent semiring in general, which makes it possible to characterize the idempotent semirings for which $\eta=\mathcal{D}^{+}$ $[\mathcal{D}^{\bullet},\mathcal{L}^{+},\mathcal{L}^{\bullet},\mathcal{R}^{+},\mathcal{R}^{\bullet}]$ in an unified and simple approach.

We present our results in two parts. The idempotent semirings $S$ such that $\eta=\mathcal{D}^{+}$ $[\mathcal{L}^{+},\mathcal{R}^{+}]$ have been characterized in [1]. The following result has some use in the last section of this articel.

Lemma 1.2**.**

[1]** Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$\mathcal{L}^{+}$* is the least distributive lattice congruence on $S$;* 2. 2.

$S\in\mathbf{LN}$* and $\mathcal{D}^{\bullet}\subseteq\mathcal{L}^{+}$;* 3. 3.

$S$* satisfies the identities:*

[TABLE] 4. 4.

$S\in\mathbf{LZ}^{+}\circ\mathbf{D}$.

In Section 2 of this article, we characterize the least distributive lattice congruence on an idempotent semiring. The idempotent semirings $S$ such that $\eta=\mathcal{D}^{\bullet}$ $[\mathcal{L}^{\bullet},\mathcal{R}^{\bullet}]$ have been studied in Section 3. In Section 4, we show that the multiplicative reduct $(S,\cdot)$ of an idempotent semiring $S$ such that $\eta=\mathcal{D}^{\bullet}$ is a normal band, and such a semiring $S$ is a spined product of an idempotent semiring $S_{1}$ such that $\eta=\mathcal{L}^{\bullet}$ and an idempotent semiring $S_{2}$ such that $\eta=\mathcal{R}^{\bullet}$ with respect to a distributive lattice.

We refer to [4] for the information concerning semigroup theory, [3] for background on semirings and [5] for notions concerning universal algebra and lattice theory.

2 The least distributive lattice congruence on an idempotent semiring

Let $S$ be an idempotent semiring. Define a binary relation $\sigma$ on $S$ by: for $a,b\in S$,

[TABLE]

Lemma 2.1**.**

Let $S$ be an idempotent semiring. Then the congruence relation generated by $\sigma$ is the least distributive lattice congruence on $S$.

Proof.

Let $a,b\in S$. Then we have $(a+b)(b+a)(a+b)+a+b+(a+b)(b+a)(a+b)=(a+b)(b+a+a+b+b+a)(a+b)=(a+b)(b+a)(a+b)$, and similarly  $(b+a)(a+b)(b+a)=(b+a)(a+b)(b+a)+a+b+(b+a)(a+b)(b+a)$, which implies that $(a+b)\sigma(b+a)$. Also $(ab)\sigma(ba)$ for all $a,b\in S$.

Now,  $a(a+ab)a+a+a(a+ab)a=(a+aba)+a+(a+aba)=a+aba=a(a+ab)a$  and  $(a+ab)a(a+ab)+a+ab+(a+ab)a(a+ab)=(a+ab)a(a+ab)$ implies that $(a+ab)\sigma a$ for all $a,b\in S$. Thus any congruence on $S$ which contains $\sigma$ is a distributive lattice congruence on $S$.

Consider a distributive lattice congruence $\rho$ on $S$ and $a,b\in S$ such that $a\sigma b$. Then $aba=aba+a+aba$ implies that $b\rho(b+aba)=(b+aba+a+aba)\rho(b+a)\rho(a+b)$. Similarly from $bab=bab+b=bab$ we have $a\rho(a+b)$ and it follows that $a\rho b$. Thus $\sigma$ is contained in every distributive lattice congruence on $S$. Hence the result follows. ∎

Now we show that $\sigma$ is not transitive on a semiring in general. For this consider the following example [2] of an idempotent semiring,

\begin{array}[]{c|ccc}+&a&b&c\\ \hline\cr a&a&b&c\\ b&b&b&b\\ c&c&b&c\end{array}                   \begin{array}[]{c|ccc}\cdot&a&b&c\\ \hline\cr a&a&a&a\\ b&b&b&b\\ c&a&b&c\end{array}

Then $a\sigma b$ and $b\sigma c$ but $a\overline{\sigma}c$ shows that $\sigma$ is not transitive.

The transitive closure $\sigma^{\ast}$ of $\sigma$ is given by:

[TABLE]

A semiring $(S,+,\cdot)$ is called almost idempotent if $(S,+)$ is a semilattice and $a+a^{2}=a^{2}$ for every $a\in S$. In [12], we proved that $\sigma^{\ast}$ is the least distributive lattice congruence on an almost idempotent semiring. Since every idempotent semiring $S$ with commutative addition is an almost idempotent semiring, $\sigma^{\ast}$ is the least distributive lattice congruence on $S$. The proof that $\sigma^{\ast}$ remains the least distributive lattice congruence on an idempotent semiring $S$ even if the addition in $S$ is not commutative, is almost similar and so we omit the proof here.

Theorem 2.2**.**

Let $S$ be an idempotent semiring. Then $\sigma^{\ast}$ is the least distributive distributive congruence on $S$.

Almost all varieties of idempotent semirings considered here are subvarieties of $\mathbf{N}$. For this, the following result will have an important roll throughout the rest of this article.

Lemma 2.3**.**

An idempotent semiring $S\in\mathbf{N}$ if and only if it satisfies the identity:

[TABLE]

Proof.

First assume that $S\in\mathbf{N}$ and $a,b,c\in S$. Then $ac=(a+abca+a)c=ac+abcac+ac=ac+abca(c+cabc+c)+ac=ac+abcac+abc+abcac+ac$ implies that $ac+abc+abcac+ac=ac$ [add $abc+abcac+ac$ to both sides from right]. Similarly this implies that $ac=ac+abc+ac$.

Converse follows directly. ∎

Theorem 2.4**.**

If an idempotent semiring $S\in\mathbf{N}$, then $\sigma$ is the least distributive lattice congruence on $S$.

Proof.

Let $a,b\in S$ be such that $a\eta b$. Then there exists $x\in S$ such that $axbxa=axbxa+a+axbxa$ and $bxaxb=bxaxb+b+bxaxb$. Since $S\in\mathbf{N}$, $ab+axb+ab=ab$ and $ba+bxa+ba=ba$, by Lemma 2.3. Then we have

[TABLE]

Similarly, $bab=bab+b+bab$. Thus $a\sigma b$ and so $\eta\subseteq\sigma$. Also $\sigma$ is contained in every distributive lattice congruence. Hence $\sigma=\eta$ is the least distributive lattice congruence on $S$. ∎

3 The idempotent semirings such that $\eta=\mathcal{D}^{\bullet},\mathcal{L}^{\bullet},\mathcal{R}^{\bullet}$

In this section we show that the class of all idempotent semirings for which $\eta=\mathcal{D}^{\bullet}[\mathcal{L}^{\bullet},\mathcal{R}^{\bullet}]$ is an equational class and so a variety. Here we find several systems of identities defining these varieties.

On an idempotent semiring $(S,+,\cdot)$ one may introduce the relations $\leq^{l}_{+},\,\leq^{l}_{\cdot},\leq^{r}_{+},\,\leq^{r}_{\cdot}$ and $\leq_{+},\,\leq_{\cdot}$ by the following: for $a,b\in S$,

[TABLE]

The relations $\leq^{l}_{+},\,\leq^{l}_{\cdot},\leq^{r}_{+},\,\leq^{r}_{\cdot}$ are quasi-orders and the relations $\leq_{+}$ and $\leq_{\cdot}$ are partial orders [6].

In [9], Pastijn and Zhao characterized the idempotent semirings $S$ such that $\mathcal{D}^{\bullet}$ is the least distributive lattice congruence on $S$. Following result is already proved in [9]. Use of Theorem 2.2 shorten the proof which we would like to include here.

Theorem 3.1**.**

Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$\mathcal{D}^{\bullet}$* is the least distributive lattice congruence on $S$;* 2. 2.

$S\in\mathbf{N}$* and $\mathcal{D}^{+}\subseteq\mathcal{D}^{\bullet}$;* 3. 3.

$S$* satisfies the identity*

[TABLE]

Proof.

$\mathbf{(1)\ \Rightarrow\ (2):}$  Let $a,b\in S$. Since $\mathcal{D}^{\bullet}$ is a distributive lattice congruence, $(a+ab)\mathcal{D}^{\bullet}a$. Then we have

[TABLE]

and so $S\in\mathbf{N}$.

Now let $a,b\in S$ such that $a\mathcal{D}^{+}b$. Then $a=a+b+a$ and $b=b+a+b$ implies that $bab=bab+b+bab$ and $aba=aba+a+aba$, and so $a\sigma b$. Thus $\mathcal{D}^{+}\subseteq\sigma\subseteq\eta=\mathcal{D}^{\bullet}$.

$\mathbf{(2)\ \Rightarrow\ (3):}$  Similar to the proof of the Theorem 2.17 [9].

$\mathbf{(3)\ \Rightarrow\ (1):}$  Let $a,b\in S$ such that $a\eta b$. Since $S$ satisfied the identity (3.1, it follows that $S\in\mathbf{N}$ and so $\eta=\sigma$, by Theorem 2.4. Hence we have

[TABLE]

Since $S$ satisfies the identity (3.1), it follows that $a=aba$ and $b=bab$. Thus $a\mathcal{D}^{\bullet}b$ and so $\eta\subseteq\mathcal{D}^{\bullet}$. Also $\mathcal{D}^{\bullet}\subseteq\sigma=\eta$. Therefore $\mathcal{D}^{\bullet}=\eta$ and so $\mathcal{D}^{\bullet}$ is the least distributive lattice congruence on $S$. ∎

Now wecharacterize the idempotent semirings for which $\mathcal{L}^{\bullet}$ is the least distributive lattice congruence. First we prove the following lemma.

Lemma 3.2**.**

Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$S$* satisfies the identity (1.1);* 2. 2.

$S\in\mathbf{N}$* and $\mathcal{R}^{\bullet}\subseteq\mathcal{D}^{+}$.*

Proof.

$\mathbf{(1)\,\Rightarrow\,(2):}$   Let $a,b\in S$ such that $a\mathcal{R}^{\bullet}b$. Then $a=ba$ and $b=ab$. Now $a=a+ab+a=a+b+a$ and $b=b+ba+b=b+a+b$ implies that $a\mathcal{D}^{+}b$. Thus $\mathcal{R}^{\bullet}\subseteq\mathcal{D}^{+}$. Also it follows trivially that $S\in\mathbf{N}$.

$\mathbf{(2)\,\Rightarrow\,(1):}$   Let $a,b\in S$. Then $ab\mathcal{R}^{\bullet}aba$ implies that $ab\mathcal{D}^{+}aba$. This implies that

[TABLE]

Thus $S$ satisfies the identity (1.1). ∎

Theorem 3.3**.**

Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$\mathcal{L}^{\bullet}$* is the least distributive lattice congruence on $S$;* 2. 2.

$\mathcal{D}^{+}\subseteq\mathcal{L}^{\bullet}$* and $S$ satisfies the identity (1.1);* 3. 3.

$S\in\mathbf{N}$* and $\mathcal{R}^{\bullet}\subseteq\mathcal{D}^{+}\subseteq\mathcal{L}^{\bullet}$;* 4. 4.

$\leq^{l}_{\cdot}\subseteq\leq_{+}$* for $S$;* 5. 5.

$S$* satisfies the identity*

[TABLE] 6. 6.

$S$* satisfies the identity*

[TABLE]

Proof.

Equivalence of (2) and (3) follows from the Lemma 3.2. Equivalence of (5) and (6) follows trivially.

$\mathbf{(1)\,\Rightarrow\,(2):}$   Let $a,b\in S$. Then $(a+ab)\mathcal{L}^{\bullet}a$ implies that $a=a(a+ab)=a+ab$ and so $a=a+ab+a$ for all $a,b\in S$. Thus $S$ satisfies the identity (1.1). Since $\mathcal{D}^{+}$ is the least semilattice congruence on the additive reduct $(S,+)$, $\mathcal{D}^{+}\subseteq\mathcal{L}^{\bullet}$.

$\mathbf{(2)\,\Rightarrow\,(5):}$   Let $a,b\in S$. Then $(a+b)\mathcal{D}^{+}(b+a)$ and so $(a+b)\mathcal{L}^{\bullet}(b+a)$. Therefore

[TABLE]

Also $S$ satisfies the identity (1.1). Thus we have

[TABLE]

for all $a,b\in S$.

$\mathbf{(5)\,\Rightarrow\,(1):}$   For every $a,b\in S$, we have $a=ab+a+ab$. This implies that $a=aba+a+aba$. Therefore $\mathcal{D}^{\bullet}$ is the least distributive lattice congruence on $S$, by Theorem 3.1. Also for every $a,b\in S$,

[TABLE]

Now let $a,b\in S$ such that $a\mathcal{D}^{\bullet}b$. Then $a=aba$ and $b=bab$. Now

[TABLE]

Similarly $b=ba$. Hence $a\mathcal{L}^{\bullet}b$ and so $\mathcal{D}^{\bullet}=\mathcal{L}^{\bullet}$. Thus $\mathcal{L}^{\bullet}$ is the least distributive lattice congruence on $S$.

$\mathbf{(4)\,\Rightarrow\,(5):}$   Let $a,b\in S$. Then $ab\leq^{l}_{\cdot}a$. This implies that $ab\leq_{+}a$. Then $ab+a=a=a+ab$ and so $a=ab+a+ab$. Thus $S$ satisfies the identity (3.2.

$\mathbf{(5)\,\Rightarrow\,(4):}$   Let $a,b\in S$ such that $a\leq^{l}_{\cdot}b$. Then $a=ba$. Also $a,b\in S$ implies that $b=ba+b+ba$. This implies that

[TABLE]

and so $a\leq_{+}b$. Therefore $\leq^{l}_{\cdot}\subseteq\leq_{+}$. ∎

The left-right dual of this result is stated as follows. Since the proof is similar to the above theorem, we omit.

Theorem 3.4**.**

Let $S$ be an idempotent semiring. Then the following conditions are equivalent.

$\mathcal{R}^{\bullet}$* is the least distributive lattice congruence on $S$;* 2. 2.

$\mathcal{D}^{+}\subseteq\mathcal{R}^{\bullet}$* and $S$ satisfies the identity (1.2);* 3. 3.

$S\in\mathbf{N}$* and $\mathcal{L}^{\bullet}\subseteq\mathcal{D}^{+}\subseteq\mathcal{R}^{\bullet}$;* 4. 4.

$\leq^{r}_{\cdot}\subseteq\leq_{+}$* for $S$;* 5. 5.

$S$* satisfies the identity*

[TABLE] 6. 6.

$S$* satisfies the identity*

[TABLE]

4 Joins and Malcev’s Products:

The variety of all idempotent semirings $S$ such that $\mathcal{D}^{\bullet}[\mathcal{L}^{\bullet},\mathcal{R}^{\bullet}]$ is the least distributive lattice congruence on $S$ will be denoted by $\mathbf{D}^{\bullet}[\mathbf{L}^{\bullet},\mathbf{R}^{\bullet}]$. Thus it follows, by Theorem 3.1, Theorem 3.3 and Theorem 3.4, that the varieties $\mathbf{D}^{\bullet},\mathbf{L}^{\bullet}$ and $\mathbf{R}^{\bullet}$ are determined by the additional identities (3.1), (3.2) and (3.4), respectively.

For subvarieties $\mathbf{V}$ and $\mathbf{W}$ of $\mathbf{I}$, the Mal’cev product $\mathbf{V}\circ\mathbf{W}$ of $\mathbf{V}$ and $\mathbf{W}$ (within $\mathbf{I}$) is the class of all idempotent semirings $S$ on which there exists a congruence $\rho$ such that $S/\rho\in\mathbf{W}$ and such that the $\rho$-classes belong to $\mathbf{V}$.

In this section, different characterizations of the varieties $\mathbf{D}^{\bullet},\mathbf{L}^{\bullet}$ and $\mathbf{R}^{\bullet}$ by Malcev’s product are given.

Theorem 4.1**.**

1. (i)

$\mathbf{L}^{\bullet}=\mathbf{LZ}^{\bullet}\circ\mathbf{D}$. 2. (ii)

$\mathbf{R}^{\bullet}=\mathbf{RZ}^{\bullet}\circ\mathbf{D}$.

Proof.

(i)  First we assume that $S\in\mathbf{LZ}^{\bullet}\circ\mathbf{D}$. Then there exists a congruence relation $\delta$ on $S$ such that $S/\delta\in\mathbf{D}$ and such that each $\delta$-class belongs to $\mathbf{LZ}^{\bullet}$. Let $a,b\in S$. Since $S/\delta\in\mathbf{D}$, we have $(a+ab)\delta a$ and $(ab+a)\delta a$. Since the $\delta$-class of $a$ is a left zero band for the multiplication, we have

[TABLE]

and so $a=(ab+a)+(a+ab)=ab+a+ab$. Thus $S$ satisfies the identity 3.2. Hence $S\in\mathbf{L}^{\bullet}$, by Theorem 3.3. Thus $\mathbf{LZ}^{\bullet}\circ\mathbf{D}\subseteq\mathbf{L}^{\bullet}$. The reverse inclusion follows trivially. Hence $\mathbf{L}^{\bullet}=\mathbf{LZ}^{\bullet}\circ\mathbf{D}$.

(ii)  The proof is left-right dual to (i). ∎

Thus for $S\in\mathbf{L}^{\bullet}$ the multiplicative reduct $(S,\cdot)$ is a left regular band and for $S\in\mathbf{R}^{\bullet}$ the multiplicative reduct $(S,\cdot)$ is a right regular band.

Lemma 4.2**.**

For an idempotent semiring $S$ the following conditions are equivalent:

$S\in\mathbf{LN}$. 2. 2.

$\mathcal{D}^{\bullet}$* is a congruence on $S$ and $S/\mathcal{D}^{\bullet}\in\mathbf{LZ}^{+}\circ\mathbf{D}$.*

Proof.

$\mathbf{(1)\,\Rightarrow\,(2):}$   Let $S\in\mathbf{LN}$. Then $S\in\mathbf{N}$ and so $\mathcal{D}^{\bullet}$ is a congruence on $S$, by Theorem 2.11 [9]. Hence $S/\mathcal{D}^{\bullet}\in\mathbf{LN}$ and $\mathcal{D}^{\bullet}_{S/\mathcal{D}^{\bullet}}\subseteq\mathcal{L}^{+}_{S/\mathcal{D}^{\bullet}}$. Thus $S/\mathcal{D}^{\bullet}\in\mathbf{LZ}^{+}\circ\mathbf{D}$.

$\mathbf{(2)\,\Rightarrow\,(1):}$   Assume that $S/\mathcal{D}^{\bullet}\in\mathbf{LZ}^{+}\circ\mathbf{D}$. Then $\mathcal{L}^{+}_{S/\mathcal{D}^{\bullet}}$ is the least distributive lattice congruence on $S/\mathcal{D}^{\bullet}$, by Lemma 1.2. Let $\overline{a}$ be the $\mathcal{D}^{\bullet}$-class of $a$ in $S$. Then for all $a,b\in S$, we have

[TABLE]

and so $S\in\mathbf{LN}$. ∎

Theorem 4.3**.**

1. (i)

$\mathbf{LN}=\mathbf{R}^{\bullet}\circ(\mathbf{LZ}^{+}\circ\mathbf{D})$. 2. (ii)

$\mathbf{RN}=\mathbf{R}^{\bullet}\circ(\mathbf{RZ}^{+}\circ\mathbf{D})$.

Proof.

(Revise)     Let $S\in\mathbf{LN}$. Then by the above lemma, $S/\mathcal{D}^{\bullet}\in\mathbf{LZ}^{+}\circ\mathbf{D}$. Hence $\mathbf{LN}\subseteq\mathbf{R}^{\bullet}\circ(\mathbf{LZ}^{+}\circ\mathbf{D})$. If $S\in\mathbf{R}^{\bullet}\circ(\mathbf{LZ}^{+}\circ\mathbf{D})$ then there exists a congruence $\rho$ on $S$ such that $S/\rho\in\mathbf{LZ}^{+}\circ\mathbf{D}$ and each $\rho$-class is in $\mathbf{R}^{\bullet}$. Hence $S/\rho\in\mathbf{LN}$, by Lemma 1.2, and so $(a+aba)\rho a$ for all $a,b\in S$. Since each $\rho$-class is in $\mathbf{R}^{\bullet}$, it follows that $a=a+aba$. Thus $\mathbf{LN}=\mathbf{R}^{\bullet}\circ(\mathbf{LZ}^{+}\circ\mathbf{D})$. ∎

Let $S_{1}$ and $S_{2}$ be two semirings and $D$ a distributive lattice. If there are two homomorphisms $\phi_{1}:S_{1}\longrightarrow D$ and $\phi_{2}:S_{2}\longrightarrow D$ onto $D$ then the semiring $S=\{(s_{1},s_{2})\in S_{1}\times S_{2}\mid\phi_{1}(s_{1})=\phi_{2}(s_{2})\}$ is called a spined product of the two semirings $S_{1}$ and $S_{2}$ with respect to the spine $D$.

We show that every semiring $S\in\mathbf{D}^{\bullet}$ is a spined product of a semiring $S_{1}\in\mathbf{L}^{\bullet}$ and a semiring $S_{2}\in\mathbf{R}^{\bullet}$. For this first we prove that the multiplicative reduct of each idempotent semiring $S\in\mathbf{D}^{\bullet}$ is a normal band. We do the groundwork by providing a sequence of useful lemmas.

Lemma 4.4**.**

Every idempotent semiring satisfies the following two identities:

[TABLE]

Proof.

Let $S$ be an idempotent semiring. Then $\mathcal{D}^{+}$ is a congruence on $S$ and $S/\mathcal{D}^{+}\in\mathcal{S}l^{+}$. Hence the multiplicative reduct $(S/\mathcal{D}^{+},\cdot)$ is a regular band [10]. Then for all $a,b,c\in S$,  $abca\mathcal{D}^{+}abaca$ and so

[TABLE]

∎

Lemma 4.5**.**

For an idempotent semiring $S$ the following conditions are equivalent:

$S\in\mathbf{D}^{\bullet}$; 2. 2.

$S$* satisfies both the identities:*

[TABLE] 3. 3.

$S$* satisfies the identity:*

[TABLE]

Proof.

It is clear that ${\textbf{(}2)\Rightarrow(3)}$ and ${\textbf{(}3)\Rightarrow(1)}$. Assume that (1) holds. Then $xz\approx(xyzx+x+xyzx)(zxyz+z+zxyz)\approx xyzxzxyz+xyzxz+xyzxzxyz+xzxyz+xz+xzxyz+xyzxzxyz+xyzxz+xyzxzxyz\approx xyz+xyzxz+xyzxzxyz+xzxyz+xz+xzxyz+xyzxzxyz+xyzxz+xyz$. This implies that $xz\approx xz+xyz$ and $xz\approx xyz+xz$. ∎

Lemma 4.6**.**

If $S\in\mathbf{D}^{\bullet}$ then $S$ satisfies the identity:

[TABLE]

Proof.

Let $S\in\mathbf{D}^{\bullet}$. Then for $x,y,z\in S$,

[TABLE]

which implies that $xyzx+xzyx\approx xyzx$ and hence $xzyx+xyzx+xzyx\approx xyzx$. ∎

Theorem 4.7**.**

For $S\in\mathbf{D}^{\bullet}$ the multiplicative reduct $(S,\cdot)$ is a normal band.

Proof.

Let $S\in\mathbf{D}^{\bullet}$ and $x,y,z,\in S$. Then $xyzx\approx xzyx+xyzx+xzyx$. Also we have

[TABLE]

∎

Theorem 4.8**.**

For every idempotent $S$ the following conditions are equivalent:

$S\in\mathbf{LNB}^{\bullet}\cap\mathbf{D}^{\bullet}$;

• 2.

$S$* satisfies the identity:*

[TABLE]

• 3.

$S\in\mathbf{L}^{\bullet}$.

Proof.

Equivalence of $(2)$ and $(3)$ is trivial.

$(1)\Rightarrow(2)$** :**   Assume that $S\in\mathbf{LNB}^{\bullet}\cap\mathbf{D}^{\bullet}$. Then we have

[TABLE]

Since $S\in\mathbf{LNB}^{\bullet}$,

[TABLE]

$(2)\Rightarrow(1)$** :**   Let $x,y,z\in S$. Then $xz\approx xzy+xz+xzy$ and $xz\approx xyz+xz+xyz$. Also we have

[TABLE]

Now $xzy\approx xzy+xzy+xzy$ implies that

[TABLE]

Thus we get $xyz\approx xzy$ and so $S\in\mathbf{LNB}^{\bullet}$. Also it follows that $S\in\mathbf{D}^{\bullet}$. ∎

Thus the multiplicative reduct $(S,\cdot)$ of every idempotent semiring $S\in\mathbf{L}^{\bullet}$ is a left normal band. Similarly, the multiplicative reduct $(S,\cdot)$ of every idempotent semiring $S\in\mathbf{R}^{\bullet}$ is a right normal band.

In $\mathbf{D}^{\bullet}$ we have the following derivation:

[TABLE]

Theorem 4.9**.**

An idempotent semiring $S\in\mathbf{D}^{\bullet}$ if and only if $S$ is a spined product of an idempotent semiring $S_{1}\in\mathbf{L}^{\bullet}$ and an idempotent semiring $S_{2}\in\mathbf{R}^{\bullet}$ with respect to a distributive lattice $D$.

Proof.

Let $S\in\mathbf{D}^{\bullet}$. Then the multiplicative reduct $(S,\cdot)$ is a regular band. So both $\mathcal{L}^{\bullet}$ and $\mathcal{R}^{\bullet}$ are congruences on the multiplicative reduct $(S,\cdot)$. Let $a,b,c\in S$ and $a\mathcal{L}^{\bullet}b$. Then $a=ab,b=ba$ and $ca\mathcal{L}^{\bullet}cb,ac\mathcal{L}^{\bullet}bc$. Hence we have,

[TABLE]

and similarly $(b+c)(a+c)=b+c$. Thus $(a+c)\mathcal{L}^{\bullet}(b+c)$. Similarly $(c+a)\mathcal{L}^{\bullet}(c+b)$. Hence $\mathcal{L}^{\bullet}$ is a congruence on $S$. Now for any $a,b\in S$,

[TABLE]

implies that $a\mathcal{L}^{\bullet}(ba+a+ba)$. Hence $S/\mathcal{L}^{\bullet}\in\mathbf{R}^{\bullet}$. Similarly $\mathcal{R}^{\bullet}$ is a congruence on $S$ and $S/\mathcal{R}^{\bullet}\in\mathbf{L}^{\bullet}$. Denote $S_{1}=S/\mathcal{L}^{\bullet}$ and $S_{2}=S/\mathcal{R}^{\bullet}$. Also $D=S/\mathcal{D}^{\bullet}$ is a distributive lattice. Since $\mathcal{L}^{\bullet}$, $\mathcal{R}^{\bullet}\subseteq\mathcal{D}^{\bullet}$, it follows that $\phi_{1}:S_{1}\longrightarrow D$ and $\phi_{2}:S_{2}\longrightarrow D$ defined by $\phi_{1}(L_{a})=D_{a}$ and $\phi_{2}(R_{a})=D_{a}$ are well defined surjective homomorphisms, where $L_{a}\,[R_{a},D_{a}]$ is the $\mathcal{L}^{\bullet}$ [ $\mathcal{R}^{\bullet}$, $\mathcal{D}^{\bullet}]$ -class containing $a$. Thus $R=\{(L_{a},R_{a})\in S_{1}\times S_{2}\mid a\mathrel{\mathcal{missing}}{D}^{\bullet}b\}$ is a spined product of $S_{1}$ and $S_{2}$ with respect to $D$.

Then the mapping $\theta:S\longrightarrow R$ defined by $\theta(a)=(L_{a},R_{a})$ is a monomorphism. Again if $(L_{a},R_{a})\in R$, then we have $a\mathrel{\mathcal{missing}}{D}^{\bullet}b$. Also since $\mathcal{D}^{\bullet}=\mathcal{L}^{\bullet}o\mathcal{R}^{\bullet}$, there exists $c\in S$ such that $a\mathrel{\mathcal{L}^{\bullet}}c$ and $c\mathrel{\mathcal{R}^{\bullet}}b$. This implies that $\theta(c)=(L_{a},R_{a})$. Thus $\theta$ is onto. ∎

Thus $\mathbf{D}^{\bullet}\subseteq\mathbf{L}^{\bullet}\vee\mathbf{R}^{\bullet}$. Also $\mathbf{L}^{\bullet}\subseteq\mathbf{D}^{\bullet}$ and $\mathbf{R}^{\bullet}\subseteq\mathbf{D}^{\bullet}$ implies that $\mathbf{L}^{\bullet}\vee\mathbf{R}^{\bullet}\subseteq\mathbf{D}^{\bullet}$. Then we have the following corollary.

Corollary 4.10**.**

$\mathbf{D}^{\bullet}=\mathbf{L}^{\bullet}\vee\mathbf{R}^{\bullet}$.

PROBLEMS

Characterize the class of all idempotent semirings $S$ such that $\sigma$ is the least distributive lattice congruence on $S$.