# Nil-extensions of simple and right $\pi$-inverse ordered semigroups

A. Jamadar
Department of Mathematics, Rampurhat College
Rampurhat-731224, West Bengal, India
<EMAIL_ADDRESS>
###### Abstract

An ordered semigroup $S$ is right $\pi$-inverse if it is $\pi$-inverse but not
conversely. So the question arises under what condition the converse holds. In
this paper we study nil-extensions of simple and right $\pi$-inverse ordered
semigroups and prove that $S$ is right $\pi$-inverse if and only if $S$ is
$\pi$-inverse in a $t$-Archimedean ordered semigroup. Moreover, we
characterize complete semilattice of nil-extensions of simple and right
$\pi$-inverse ordered semigroups.

Key Words and phrases: $l$-Archimedean, $t$-Archimedean, $\pi$-regular, nil-
extension, ordered idempotent, simple ordered semigroup, right $\pi$-inverse
ordered semigroup.
2010 Mathematics subject Classification: 20M10; 06F05.

##  1 Introduction and Preliminaries

An ordered semigroup is a partiality ordered set $(S,\leq)$ and at the same
time a semigroup $(S,\cdot)$ such that for all $a,\;b,\;x\in S,$ $a\leq
b\;\textrm{implies}\;xa\leq xb\;\textrm{and}\;ax\leq bx$. It is denoted by
$(S,\cdot,\leq)$. For a subset $A$ of $S$, let $(A]=\\{x\in S:x\leq
a,\;\textrm{for some}\;a\in A\\}$.

A nonempty subset $A$ of $S$ is called a left (right) ideal [8] of $S$, if
$SA\subseteq A\;(AS\subseteq A)$ and $(A]=A$. A nonempty subset $A$ is called
a (two-sided) ideal of $S$ if it is both a left and a right ideal of $S$. An
left (right) ideal $I$ of $S$ is proper if $I\neq S$. $S$ is left(right)
simple if it does not contain proper left (right) ideals. An ordered semigroup
$S$ is called simple if for every ideal $I$ of $S$, we have $I=S$. $S$ is
called t-simple if it is both left and right simple.

The principal [8] left ideal, right ideal, ideal and bi-ideal generated by
$a\in S$ are denoted by $L(a),\;R(a),\;I(a)$and $B(a)$ respectively and
defined by

$L(a)=(a\cup Sa],\;R(a)=(a\cup aS],\;I(a)=(a\cup Sa\cup aS\cup
SaS]\;and\;B(a)=(a\cup aSa].$

Kehayopulu [8] defined Greens relations
$\mathcal{L},\;\mathcal{R},\;\mathcal{J}\;\textrm{and}\;\mathcal{H}$ on an
ordered semigroup $S$ as follows:

$a\mathcal{L}b\;if\;L(a)=L(b),\;a\mathcal{R}b\;if\;R(a)=R(b),\;a\mathcal{J}b\;if\;I(a)=I(b)\;\textrm{and}\;\mathcal{H}=\;\mathcal{L}\cap\;\mathcal{R}.$

These four relations are equivalence relations on $S$.

An element $a$ of $S$ is said to be regular if $a\in(aSa]$. We denote set of
regular elements by $Reg_{\leq}(S)$. An element $b\in S$ is ordered inverse
[3] of $a$ if $a\leq aba$ and $b\leq bab$. The set of all ordered inverses of
an element $a\in S$ is denoted by $V_{\leq}(a)$. An element $e\in S$ is said
to be ordered idempotent if $e\leq e^{2}$. The set of all ordered idempotents
of $S$ is denoted by $E_{\leq}(S)$.

An ordered semigroup $S$ is called Archimedean [2] if for every $a,b\in S$
there is $m\in\mathbb{N}$ such that $b^{m}\in(SaS]$. $S$ is called
$r(l\;or\;t)$-Archimedean [2] if for every $a,b\in S$, there exists
$m\in\mathbb{N}$ such that
$b^{m}\in(aS]\;(b^{m}\in(Sa]\;\textrm{or}\;b^{m}\in(aSa])$.

An ordered semigroup $S$ is called $\pi$-regular (resp. intra $\pi$-regular)
[2] if for every $a\in S$ there is $m\in\mathbb{N}$ such that
$a^{m}\in(a^{m}Sa^{m}]$ (resp. $a^{m}\in(Sa^{2m}S]$). We denote set of all
$\pi$-regular and intra $\pi$-regular elements by $\pi Reg_{\leq}(S)$ and $\Pi
Intra_{\leq}(S)$ respectively. A $\pi$-regular ordered semigroup $S$ is called
right $\pi$-inverse [6] if for every $a\in S$, there is $m\in\mathbb{N}$ such
that any two inverses of $a^{m}$ are $\mathcal{R}$-related.

Nil-extensions of an ordered semigroup $S$ with zero $0$ are precisely the
ideal extensions of an ideal $I$ of $S$ by the nilpotent ordered semigroup
$S/I$ [9]. Following Jamadar [7], a right $\pi$-inverse ordered semigroup is a
$\pi$-regular semigroup with the property that each element $a\in S$ there
exists $m\in\mathbb{N}$ such that any two inverses of $a^{m}$ are
$\mathcal{R}$-related. These ordered semigroups are natural generalization of
$\pi$-inverse ordered semigroups. Hansda and Jamadar [4] studied ordered
semigroups which are nil-extensions of both simple and $\pi$-inverse ordered
semigroups. So it is a logical step to study ordered semigroups which are nil-
extensions of both simple and right $\pi$-inverse ordered semigroups. Aim of
this work is to describe nil-extensions of simple and right $\pi$-inverse
ordered semigroups. Furthermore complete semilattice decompositions of the
nil-extensions of simple and right $\pi$-inverse ordered semigroups have been
given here. This paper is inspired by [1] and [4].

A congruence $\rho$ on $S$ is called semilattice if for all $a,b\in
S\;a\;\rho\;a^{2}\;\textrm{and}\;ab\;\rho\;ba$. A semilattice congruence
$\rho$ on $S$ is called complete if $a\leq b$ implies $(a,ab)\in\rho$. The
ordered semigroup $S$ is called complete semilattice of subsemigroup of type
$\tau$ if there exists a complete semilattice congruence $\rho$ such that
$(x)_{\rho}$ is a type $\tau$ subsemigroup of $S$. Equivalently: There exists
a semilattice $Y$ and a family of subsemigroups $\\{S\\}_{\alpha\in Y}$ of
type $\tau$ of $S$ such that:

  1. 1.

$S_{\alpha}\cap S_{\beta}=\;\phi$ for any
$\alpha,\;\beta\in\;Y\;with\;\alpha\neq\beta,$

  2. 2.

$S=\bigcup_{\alpha\;\in\;Y}\;S_{\alpha},$

  3. 3.

$S_{\alpha}S_{\beta}\;\subseteq\;S_{\alpha\;\beta}$ for any
$\alpha,\;\beta\in\;Y,$

  4. 4.

$S_{\beta}\cap(S_{\alpha}]\neq\phi$ implies $\beta\;\preceq\;\alpha,$ where
$\preceq$ is the order of the semilattice $Y$ defined by
$\preceq:=\\{(\alpha,\;\beta)\;\mid\;\alpha=\alpha\;\beta\;(\beta\;\alpha)\\}$
[10].

Let $S$ be a $\pi$-regular ordered semigroup. Due to Sadhya and Hansda [11],
the following equivalence relations $\mathcal{L}^{*}$, $\mathcal{R}^{*}$,
$\mathcal{J}^{*}$ and $\mathcal{H}^{*}$ are given by:

  1. 1.

$a\mathcal{L}^{*}b\Leftrightarrow a^{m}\mathcal{L}b^{n}$

  2. 2.

$a\mathcal{R}^{*}b\Leftrightarrow a^{m}\mathcal{R}b^{n}$

  3. 3.

$a\mathcal{J}^{*}b\Leftrightarrow a^{m}\mathcal{J}b^{n}$

  4. 4.

$\mathcal{H}^{*}=\mathcal{L}^{*}\cap\mathcal{R}^{*}$

where $a,b\in S$ and $m,n$ are the smallest positive integers such that
$a^{m},b^{n}\in Reg_{\leq}(S)$.

For $a,b\in S$, $a|b$ if and only if there exists $x,y\in S^{1}$ such that
$b\leq xay$.

######  Theorem 1.1.

[7] A $\pi$-regular ordered semigroup $S$ is a right $\pi$-inverse if and only
if for any two idempotents $e,f\in E_{\leq}(S)$, $e\mathcal{L}^{*}f$ implies
$e\mathcal{R}^{*}f$.

######  Theorem 1.2.

[7] The following conditions are equivalent on a $\pi$-regular ordered
semigroup $S$.

  1. 1.

$S$ is right $\pi$-inverse;

  2. 2.

for $a\in S$ there is $m\in\mathbb{N}$ such that
$a^{\prime},a^{\prime\prime}\in V_{\leq}(a^{m})$ implies
$a^{\prime}\mathcal{R}a^{\prime\prime}$.

######  Theorem 1.3.

[4] The following conditions on an ordered semigroup $S$ are equivalent:

  * (i)

$S$ is a nil-extension of a left simple and $\pi$-inverse ordered semigroup;

  * (ii)

$S$ is $\pi$-inverse and $l$-Archimedean ordered semigroup;

  * (iii)

$S$ is $\pi$-inverse and $a\mathcal{L}^{*}b$ for every $a,b\in S$;

  * (iv)

$S$ is $\pi$-inverse and $e\mathcal{L}^{*}f$ for every $e,f\in E_{\leq}(S)$;

  * (v)

$S$ is $\pi$-regular and $e\mathcal{H}^{*}f$ for every $e,f\in E_{\leq}(S)$;

  * (vi)

$S$ is $\pi$-regular and $a\mathcal{H}^{*}b$ for every $a,b\in S$;

  * (vii)

$S$ is $\pi$-inverse and $t$-Archimedean ordered semigroup;

  * (viii)

$S$ is a nil-extension of $t$-simple and $\pi$-inverse ordered semigroup.

######  Corollary 1.4.

[4] The following conditions on an ordered semigroup $S$ are equivalent:

  * (i)

$S$ is a nil-extension of a simple and $\pi$-inverse ordered semigroup;

  * (ii)

$S$ is $\pi$-inverse and $e\mathcal{J}^{*}f$ for all $e,f\in E_{\leq}(S)$.

######  Lemma 1.5.

[2] Let $S$ be an ordered semigroup and $I$ an ideal of $S$. Then the
following conditions are equivalent:

  * (i)

$S$ is a nil-extension of $I$;

  * (ii)

$(\forall a\in S)(\exists m\in\mathbb{N})\;a^{m}\in I$.

######  Corollary 1.6.

[2] The following conditions are equivalent on a poe-semigroup $S$:

  * (i)

$S$ is a nil-extension of a $t$-simple po-semigroup;

  * (ii)

$S$ is a $t$-Archimedean po-semigroup in which $Intra(S)\neq\phi$.

##  2 Nil-extensions of simple and right $\pi$-inverse ordered semigroups

In this section we describe all ordered semigroups which are the nil-
extensions of simple and right $\pi$-inverse, left simple and right
$\pi$-inverse ordered semigroups. We define the sets $\textbf{R}{V}_{\leq}(S)$
and $\Pi\mathbf{R}{V}_{\leq}(S)$ as follows:

$\textbf{R}{V}_{\leq}(S)=\\{a\in S\;\mid\;\textrm{for}\;\textrm{any}\;x,\;y\in
V_{\leq}(a)\;\textrm{implies}\;x\mathcal{R}y\;\\},$

$\Pi\mathbf{R}{V}_{\leq}(S)=\\{a\in S\;(\exists
m\in\mathbb{N})\;\mid\;\textrm{for}\;\textrm{any}\;x,\;y\in
V_{\leq}(a^{m})\;\textrm{implies}\;x\mathcal{R}y\;\\}.$

######  Lemma 2.1.

Let $S$ be an ordered semigroup. Then the following conditions are equivalent
on $S$:

  * (i)

for all $a\in S$ and for all $c\in\mathbf{R}V_{\leq}(S)$, $a\mid c$ implies
$a^{2}\mid c$;

  * (ii)

for all $a,b\in S$ and for all $c\in\mathbf{R}V_{\leq}(S)$, $a\mid
c\;and\;b\mid c$ implies $ab\mid c$.

###### Proof.

This follows from [4].

∎

######  Theorem 2.2.

Let an ordered semigroup $S$ be a complete semilattice $Y$ of subsemigroups
$\\{S_{\alpha}\\}_{\alpha\in Y}$. Then

  * (i)

$\textbf{R}{V}_{\leq}(S)=\cup\textbf{R}{V}_{\leq}(S_{\alpha})$.

  * (ii)

$S$ is right inverse if and only if $S_{\alpha}$ is right inverse for all
$\alpha\in Y$.

  * (iii)

$S$ is right $\pi$-inverse if and only if $S_{\alpha}$ is right $\pi$-inverse
for all $\alpha\in Y$.

###### Proof.

This follows from [4]. ∎

######  Lemma 2.3.

Let an ordered semigroup $S$ be a nil-extension of a semigroup $K$ of type
$\tau$ and $\mathbf{V}_{\leq}(S)\neq\phi$. Then the following statements hold
in $S$.

  * (i)

For every $a\in\mathbf{R}V_{\leq}(S),\;a\in K$.

  * (ii)

For every $\mathcal{L}$-class of $S$ that contains an
$a\in\mathbf{R}V_{\leq}(S)$ is a subset of $K$.

###### Proof.

This follows from [4].

∎

Following [6] we know that if $S$ is a $\pi$-inverse ordered semigroup then it
is right $\pi$-inverse but not conversely. In the next theorem we describe
ordered semigroups which are nil-extensions of simple and right $\pi$-inverse
ordered semigroups and show that $S$ is $\pi$-inverse if and only if it is
right $\pi$-inverse in a $t$-Archimedean ordered semigroup. Also we prove that
in a right $\pi$-inverse ordered semigroup, $S$ is $l$-Archimedean if and only
if it is $t$-Archimedean.

######  Theorem 2.4.

The following conditions on an ordered semigroup $S$ are equivalent:

  * (i)

$S$ is a nil-extension of a left simple and right $\pi$-inverse ordered
semigroup;

  * (ii)

$S$ is right $\pi$-inverse and $l$-Archimedean ordered semigroup;

  * (iii)

$S$ is right $\pi$-inverse and $a\mathcal{L}^{*}b$ for any $a,b\in S$;

  * (iv)

$S$ is right $\pi$-inverse and $e\mathcal{L}^{*}f$ for any $e,f\in
E_{\leq}(S)$;

  * (v)

$S$ is $\pi$-regular and $e\mathcal{H}^{*}f$ for any $e,f\in E_{\leq}(S)$;

  * (vi)

$S$ is $\pi$-regular and $a\mathcal{H}^{*}b$ for any $a,b\in S$;

  * (vii)

$S$ is $\pi$-inverse and $t$-Archimedean ordered semigroup;

  * (viii)

$S$ is right $\pi$-inverse and $t$-Archimedean ordered semigroup;

  * (ix)

$S$ is a nil-extension of a $t$-simple and right $\pi$-inverse ordered
semigroup.

###### Proof.

$(i)\Rightarrow(ii)$: Let $S$ be a nil-extension of a left simple and right
$\pi$-inverse ordered semigroup $K$. Choose $a\in S$. Then there exists
$k\in\mathbb{N}$ such that $a^{k}\in K$. Since $K$ is right $\pi$-inverse, for
$a^{k}$ there exists $r\in\mathbb{N}$ such that for any $x,y\in
V_{\leq}(a^{m})\subseteq K$ implies $x\mathcal{R}y$, where $m=kr$, by Theorem
1.2. Hence $S$ is right $\pi$-inverse ordered semigroup. Also $S$ is
$l$-Archimedean by Theorem 1.3.

$(ii)\Rightarrow(iii)$ and $(iii)\Rightarrow(iv)$: These are obvious.

$(iv)\Rightarrow(v)$: Since $S$ is right $\pi$-inverse so $S$ is $\pi$-regular
and $e\mathcal{L}^{*}f$ implies $e\mathcal{H}^{*}f$ for any $e,f\in
E_{\leq}(S)$, by Theorem 1.1. Hence $e\mathcal{H}^{*}f$.

$(v)\Rightarrow(vi)$, $(vi)\Rightarrow(vii)$, $(vii)\Rightarrow(viii)$ and
$(viii)\Rightarrow(ix)$: This follows from Theorem 1.3.

$(ix)\Rightarrow(i)$: Since $S$ is $t$-Archimedean so $S$ is a nil-extension
of a $t$-simple ordered semigroup $K$ by Corollary 1.6. So $K$ is a left
simple ordered semigroup. Let $a\in K$. Since $S$ is right $\pi$-inverse, for
$a\in S$ there exists $m\in\mathbb{N}$ such that for any
$a^{\prime},a^{\prime\prime}\in V_{\leq}(a^{m})$ implies
$a^{\prime}\mathcal{R}a^{\prime\prime}$. Now since $K$ is an ideal, so
$a^{\prime}a^{m}a^{\prime}\in K$. Hence $a^{\prime}\leq
a^{\prime}a^{m}a^{\prime}$ implies $a^{\prime}\in K$. Similarly
$a^{\prime\prime}\in K$. So $a^{\prime}\mathcal{R}a^{\prime\prime}$ in $K$.
Hence $K$ is a right $\pi$-inverse ordered semigroup.

∎

######  Corollary 2.5.

The following conditions on an ordered semigroup $S$ are equivalent:

  * (i)

$S$ is a nil-extension of a simple and right $\pi$-inverse ordered semigroup;

  * (ii)

$S$ is right $\pi$-inverse and $e\mathcal{J}^{*}f$ for all $e,f\in
E_{\leq}(S)$;

  * (iii)

$S$ is right $\pi$-inverse and $a\mathcal{J}^{*}b$ for all $a,b\in S$;

  * (iv)

$S$ is right $\pi$-inverse and for all $a,b\in S$, there exists
$m\in\mathbb{N}$ such that $a^{m}\in(SbS]$;

  * (v)

$S$ is right $\pi$-inverse and Archimedean ordered semigroup.

###### Proof.

$(i)\Rightarrow(ii)$: Let $S$ be a nil-extension of a simple and right
$\pi$-inverse ordered semigroup $K$. Choose $a\in S$. Then there exists
$k\in\mathbb{N}$ such that $a^{k}\in K$. Since $K$ is right $\pi$-inverse, for
$a^{k}$ there exists $r\in\mathbb{N}$ and for any $x,y\in K$ such that $x,y\in
V_{\leq}(a^{m})$ implies $x\mathcal{R}y$, where $m=kr$. Hence $S$ is right
$\pi$-inverse ordered semigroup. Also $e\mathcal{J}^{*}f$ for all $e,f\in
E_{\leq}(S)$, by Corollary 1.4.

$(ii)\Rightarrow(iii)$, $(iii)\Rightarrow(iv)$ and $(iv)\Rightarrow(v)$: These
are obvious.

$(v)\Rightarrow(i)$: Since $S$ is an Archimedean ordered semigroup, so $S$ is
nil-extension of the simple ordered semigroup $K$ by Theorem $3.3$ of [2]. Let
$a\in K$. Since $S$ is right $\pi$-inverse, for $a\in S$ there exists
$m\in\mathbb{N}$ such that for any $a^{\prime},a^{\prime\prime}\in
V_{\leq}(a^{m})$ implies $a^{\prime}\mathcal{R}a^{\prime\prime}$. Now since
$K$ is an ideal, so $a^{\prime}a^{m}a^{\prime}\in K$. Hence $a^{\prime}\leq
a^{\prime}a^{m}a^{\prime}$ implies $a^{\prime}\in K$. Similarly
$a^{\prime\prime}\in K$. So $a^{\prime}\mathcal{R}a^{\prime\prime}$ in $K$.
Hence $K$ is a right $\pi$-inverse ordered semigroup

∎

######  Corollary 2.6.

An ordered semigroup $S$ is a nil-extension of a right inverse ordered
semigroup if and only if the following conditions hold in $S$:

  * (i)

$S$ is right $\pi$-inverse;

  * (ii)

for $a\in S$ and $b\in\mathbf{R}V_{\leq}(S)$ such that $a\leq ba$ implies that
$a\in\mathbf{R}V_{\leq}(S)$;

  * (iii)

for $a\in S$ and $b\in\mathbf{R}V_{\leq}(S)$ such that $a\leq ab$ implies that
$a\in\mathbf{R}V_{\leq}(S)$;

  * (iv)

for $a\in S$ and $b\in\mathbf{R}V_{\leq}(S)$ such that $a\leq b$ implies that
$a\in\mathbf{R}V_{\leq}(S)$.

###### Proof.

First suppose that $S$ is a nil-extension of a right inverse ordered semigroup
$K$.

$(i)$. Let $a\in S$. Then there is $m\in\mathbb{N}$ such that $a^{m}\in K$.
Since $K$ is right inverse, so there are $x,y\in K$ such that for any $x,y\in
V_{\leq}(a^{m})$ implies $x\mathcal{R}y$. Thus $S$ is right $\pi$-inverse.

$(ii)$. Let $b\in\mathbf{R}V_{\leq}(S)$ and $a\in S$ such that $a\leq ba$.
Then since $b\in Reg_{\leq}(S),\\\ \;\textrm{there is}\;z\in S$ such that
$b\leq b(zb)^{n}\;\textrm{for all}\;n\in\mathbb{N}$. Let $n_{1}\in\mathbb{N}$
such that $(zb)^{n_{1}}\in K$. Then $b(zb)^{n_{1}}\in K$, which implies $b\in
K$ and so $ba\in K$ and finally $a\in K$. Since $K$ is a right inverse ordered
semigroup, $a\in\mathbf{R}V_{\leq}(S)$.

$(iii)$. This is similar to $(ii)$.

$(iv)$. Let $a\in S$ and $b\in\mathbf{R}V_{\leq}(S)$ such that $a\leq b$.
Clearly $b\in Reg_{\leq}(S)$, and so for some $z\in S,\;b\leq b(zb)^{n}$,
which holds for all $n\in\mathbb{N}$. Since $S$ is nil-extension of $K$ there
is $m\in\mathbb{N}$ such that $(zb)^{m}\in K$ and so $b\in K$. Thus $a\in K$.
Therefore $a\in\mathbf{R}V_{\leq}(S)$.

Conversely, assume that given conditions hold in $S$. Let $a\in S$ be
arbitrary. Then by (i) there exists $m\in\mathbb{N}$ such that for any $x,y\in
S$, $x,y\in V_{\leq}(a^{m})$ implies $x\mathcal{R}y$. So
$\mathbf{R}V_{\leq}(S)\neq\phi$. Say $T=\mathbf{R}V_{\leq}(S)$. Thus for each
$a\in S,\;\textrm{there exists}\;m\in\mathbb{N}$ such that for any $x,y\in S$,
$x,y\in V_{\leq}(a^{m})$ implies $x\mathcal{R}y$. Now choose $s\in S$ and
$a\in T$. Then $a\in Reg_{\leq}(S)$, there is $h\in S$ such that $a\leq
a(ha)^{n}\;\textrm{for all}\;n\in\mathbb{N}$. Let $m_{1}\in\mathbb{N}$ such
that $(ha)^{m_{1}}\in T$. So $sa\leq sa(ha)^{m_{1}}$ implies that $sa\in
V_{\leq}(S)=T$, by (iii). Similarly $as\in T$ follows from (ii).

Now consider $a\in S$ and $b\in T$ such that $a\leq b$. Then by (iv) $a\in T$.
Also $T$ is a right inverse ordered semigroup. Hence $S$ is nil-extension of a
right inverse ordered semigroup $T$. ∎

##  3 Complete semilattice of nil-extensions of right $\pi$-inverse ordered
semigroups

In this section we characterize complete semilattice decompositions of nil-
extensions of right $\pi$-inverse ordered semigroups.

######  Corollary 3.1.

Let $S$ be an ordered semigroup. Then the following conditions are equivalent
on $S$:

  * (i)

$S$ is a complete semilattice of nil-extensions of simple and right
$\pi$-inverse ordered semigroups;

  * (ii)

$S$ is a complete semilattice of nil-extensions of simple semigroups and $\Pi
Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$;

  * (iii)

$S$ is right $\pi$-inverse and is a complete semilattice of Archimedean
ordered semigroups.

###### Proof.

$(i)\Rightarrow(ii)$: Clearly $S$ is a complete semilattice of semigroups
$\\{S_{\alpha}\\}_{\alpha\in Y}$, where $S_{\alpha}$ is a nil-extension of
simple ordered semigroup $K_{\alpha}$ and $\rho$ be the corresponding complete
semilattice congruence on $S$. So we need only to show that $\Pi
Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$. For this, let
$a\in\Pi\mathbf{R}{V}_{\leq}(S)$. Then clearly there are
$m\in\mathbb{N}\;\textrm{and}\;\alpha\in Y$ such that $a^{m}\in K_{\alpha}$
and $a^{m}\in(K_{\alpha}a^{2m}K_{\alpha}]$. Thus $a\in\Pi Intra_{\leq}(S)$.

Also let $b\in\Pi Intra_{\leq}(S)$. Then there are $x,y\in S$ and $\gamma\in
Y$ such that $b^{m}\leq xb^{2m}y$ and $b\in S_{\gamma}$. Let $S_{\gamma}$ is a
nil-extension of simple and right $\pi$-inverse ordered semigroup
$K_{\gamma}$. Now $b^{m}\leq(xb^{m})^{n}b^{m}y^{n}$ for all $n\in\mathbb{N}$.
Also by completeness of $\rho$ we have that
$(b^{m})_{\rho}=(b^{m}xb^{2m}y)_{\rho}=(xb^{m}xb^{m}y)_{\rho}=(xb^{m})_{\rho}(xb^{m}y)_{\rho}=(xb^{m})_{\rho}(b^{m})_{\rho}=(xb^{m})_{\rho}.$
Thus $xb^{m}\in S_{\gamma}$. So there is $m_{1}\in\mathbb{N}$ such that
$(xb^{m})^{m_{1}}\in K_{\gamma}$. Thus $b^{m_{1}}\in K_{\gamma}$. Since
$K_{\gamma}$ is right $\pi$-inverse it follows that
$b\in\Pi\mathbf{R}{V}_{\leq}(S)$. Hence $\Pi
Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$.

$(ii)\Rightarrow(iii)$: Suppose $S$ is a complete semilattice of semigroups
$S_{\alpha}(\alpha\in Y)$, where $S_{\alpha}$ is a nil-extension of simple
ordered semigroup $K_{\alpha}$ and $\Pi
Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$. Let $a\in S$. Then there are
$m\in\mathbb{N}$ and $\alpha\in Y$ such that $a^{m}\in K_{\alpha}$. Since each
$K_{\alpha}$ is simple, for $a^{m}\in K_{\alpha}$ there exists
$r\in\mathbb{N}$ such that $(a^{m})^{r}\in(K_{\alpha}(a^{m})^{2r}K_{\alpha}]$.
Thus $a\in\Pi Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$. Hence $S$ is right
$\pi$-inverse ordered semigroup. Also $S$ is a complete semilattice of
Archimedean ordered semigroups, by Corollary 3.9 of [2].

$(iii)\Rightarrow(i)$: Suppose that the condition (iii) holds. Then $S$ is a
complete semilattice of nil-extensions of simple semigroups, by Theorem $3.3$
of [2]. Suppose that $S$ is a complete semilattice of semigroups
$S_{\alpha}(\alpha\in Y)$, where $S_{\alpha}$ is nil-extension of simple
semigroup $K_{\alpha}$. Let $a\in K_{\alpha}$. Since $S$ is $\pi$-inverse,
there is $m\in\mathbb{N}$ such that for any $z,y\in V_{\leq}(a^{m})$ implies
that $z\mathcal{R}y$, where $z,y\in S$. Now $a^{m}\leq a^{m}za^{m}$ implies by
completeness of $\rho$, $(a^{m})_{\rho}=(za^{m})_{\rho}$ and so
$a^{m},za^{m}\in S_{\alpha}$. Now $a^{m}\leq a^{m}za^{m}$ implies that
$a^{m}\leq a^{m}(za^{m}z)a^{m}$. Since $S_{\alpha}$ is a nil-extension of
$K_{\alpha}$, $K_{\alpha}$ is an ideal of $S_{\alpha}$. Thus $za^{m}z\in
K_{\alpha}$. Now $z\leq za^{m}z$ implies by completeness of $\rho$,
$(z)_{\rho}=(za^{m}z)_{\rho}$. So $z\in K_{\alpha}$. Similarly $y\in
K_{\alpha}$. Hence $K_{\alpha}$ is right $\pi$-inverse and so $S$ is complete
semilattice of nil-extensions simple and right $\pi$-inverse ordered
semigroups.

∎

######  Corollary 3.2.

Let $S$ be an ordered semigroup. Then the following conditions are equivalent
on $S$:

  * (i)

$S$ is a complete semilattice of nil-extensions of left simple and right
$\pi$-inverse ordered semigroups;

  * (ii)

$S$ is a complete semilattice of nil-extensions of left simple semigroups and
$\Pi Intra_{\leq}(S)=\Pi\mathbf{R}{V}_{\leq}(S)$;

  * (iii)

$S$ is right $\pi$-inverse and is a complete semilattice of $l$-Archimedean
ordered semigroups.

###### Proof.

This follows from Theorem 2.4 and Corollary 3.1.

∎

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