# The Natural partial order on modules

T. Pakel  Tugba Pakel, Department of Mathematics, Ankara University, Turkey
<EMAIL_ADDRESS>,  T. P. Calci  Tugce Pekacar Calci, Department of
Mathematics, Ankara University, Turkey<EMAIL_ADDRESS>,  S. Halicioglu
Sait Halıcıoglu, Department of Mathematics, Ankara University, Turkey
<EMAIL_ADDRESS>,  A. Harmanci  Abdullah Harmanci, Department of
Mathematics, Hacettepe University, Turkey<EMAIL_ADDRESS>and  B.
Ungor  Burcu Ungor, Department of Mathematics, Ankara University, Turkey
<EMAIL_ADDRESS>
###### Abstract.

The Mitsch order is already known as a natural partial order for semigroups
and rings. The purpose of this paper is to further study of the Mitsch order
on modules by investigating basic properties via endomorphism rings. And so
this study also contribute to the results related to the orders on rings. As a
module theoretic analog of the Mitsch order, we show that this order is a
partial order on arbitrary modules. Among others, lattice properties of the
Mitsch order and the relations between the Mitsch order and the other well-
known orders, such as, the minus order, the Jones order, the direct sum order
and the space pre-order on modules are studied. In particular, we prove that
the minus order is the Mitsch order and we supply an example to show that the
converse does not hold in general.

2010 MSC: 06A06, 06F25, 06F99, 16B99

Key words: Mitsch order, minus order, Jones order, direct sum order, space
pre-order

##  1\. Introduction

Throughout this paper $R$ denotes an associative ring with identity $1_{R}$
and modules are unitary right $R$-modules. For a module $M$, $S=$
End${}_{R}(M)$ stands for the ring of all right $R$-module endomorphisms of
$M$. It is well known that $M$ is a left $S$-right $R$-bimodule. In this work,
for the $(S,R)$-bimodule $M$, $l_{S}(.)$ and $r_{R}(.)$ stand for the left
annihilator of a subset of $M$ in $S$ and the right annihilator of a subset of
$M$ in $R$, respectively. If the subset is a singleton, say $\\{m\\}$, then we
simply write $l_{S}(m)$ and $r_{R}(m)$, respectively. For a ring $R$, $C(R)$
denotes the center of $R$.

Let $\mathcal{S}$ be a semigroup and $a\in\mathcal{S}$. Any solution $x=a^{-}$
to the equation $axa=a$ is called an inner generalized inverse of $a$. If in
addition $xax=x$, then $x$ is called a reflexive inverse of $a$ and denoted by
$a^{+}$. If such $a^{-}$ exists, then $a$ is called regular, and if every
element in a semigroup $\mathcal{S}$ is regular, then $\mathcal{S}$ is called
a  regular semigroup. Hartwig [6] introduced the plus order $\leq^{+}$ on
regular semigroups using generalized inverses. For a regular semigroup
$\mathcal{S}$ and $a,b\in\mathcal{S}$, we write

$a\leq^{+}b\quad\text{if\quad}a^{+}a=a^{+}b\quad\text{and\quad}aa^{+}=ba^{+}$
(1)

for some reflexive inverse $a^{+}$ of $a$. It is shown that the plus order is
a partial order. On the other hand, consider

$a\leq^{-}b\quad\text{if\quad}a^{-}a=a^{-}b\quad\text{and\quad}aa^{-}=ba^{-}$
(2)

for some inner generalized inverse $a^{-}$ of $a$. The relation $\leq^{-}$ is
usually called minus order. It is well known that definitions (1) and (2) are
equivalent. In [7], the plus partial order was renamed as minus partial order.

Let $S$ be any semigroup and $S^{1}$ denote the set $S$ if $S$ has an
identity. In [12], the Mitsch order was introduced for semigroups by Mitsch as
follows: for $a,b\in S$,

$a\leq b$ if $a=xb=by$, $xa=a$

for some $x,y\in S^{1}$. Then it is a partial order on $S$ which is called the
Mitsch order.

Principally projective rings were introduced by Hattori [8] to study the
torsion theory, that is, a ring is called left (resp. right) principally
projective if every principal left (resp. right) ideal is projective. This is
equivalent to the left (resp. right) annihilator of any element of the ring is
generated by an idempotent as a left (resp. right) ideal, i.e., the ring is
left (right) Rickart. If a ring $R$ is both left and right Rickart, then it is
said to be Rickart.

Some partial orders are defined on Baer and Rickart rings in the literature
(see [3], [10] and [14]). Let $R$ be a ring and $a,b\in R$. Then we write
$a\leq^{-}b$ if there exist idempotent elements $p,q\in R$ such that

$l_{R}(a)=R(1_{R}-p),\quad r_{R}(a)=(1_{R}-q)R,\quad
pa=pb,\quad\text{and}\quad aq=bq.$ (3)

It was proved in [3] that this relation $\leq^{-}$ is indeed a partial order
when $R$ is a Rickart ring. In [5], the Mitsch order defined in a ring in the
following way: For any ring $R$, a natural partial order $\mathcal{M}$ on $R$
defined by $x\mathcal{M}y$ for given $x,y\in R$ if either $x=y$ or there exist
$p,q\in R$ such that $px=py=x=xq=yq$. In [4], it is shown that the minus order
and the Mitsch order are the same on Rickart rings.

Let $M$ be a module. Then an element $m\in M$ is called (Zelmanowitz) regular
if $m=m\varphi(m)\equiv\hskip 2.84544ptm\varphi m$ for some $\varphi\in M^{*}$
where $M^{*}=$ Hom${}_{R}(M,R)$ denotes the dual of $M$. For a ring $R$, let
$a\in R$ be a regular element (in the sense of von Neumann), i.e., there
exists $a^{-}\in R$ such that $a=aa^{-}a$. Define the map
$\varphi:R\rightarrow R$ with $\varphi(r)=a^{-}r,r\in R$. Then $\varphi\in
R^{*}=$ End${}_{R}(R)$. We have $a\varphi a=a\varphi(a)=aa^{-}a=a$ which
yields that $a$ is regular in $R_{R}$ (in the sense of Zelmanowitz, see also
[17]). Let now $R$ be a ring and suppose $a$ is a regular element in $R_{R}$
(in the sense of Zelmanowitz), i.e., there exists $\varphi\in R^{*}$ such that
$a=a\varphi a$. Define $a^{-}=\varphi(1_{R})$. Then $a=a\varphi
a=a\varphi(1_{R})a=a\varphi(1_{R}a)=a\varphi(a)=a\varphi a=a.$ We may conclude
that $a\in R$ is regular if and only if $a$ is regular in $R_{R}$ (or,
similarly, in the left-$R$ module ${}_{R}R$). A module $M$ is called regular
(in the sense of Zelmanowitz) if every element of $M$ is regular.

Recently, some of the present authors study the minus order in a more general
setting. The notion of the minus order was extended to modules using their
endomorphism rings in [15]. Let $M$ be a module and $m_{1}$, $m_{2}\in M$.
Then $m_{1}\leq^{-}m_{2}$ if there exists $\varphi\in M^{*}$ such that
$m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$ and $\varphi
m_{1}=\varphi m_{2}$. They called the relation $\leq^{-}$ the minus order on
$M$, and proved that it is a partial order on regular modules. In the same
work, the authors also defined the Mitsch order on modules, that is,
$m_{1}\leq_{M}m_{2}$ if there exist $f\in S$ and $a\in R$ such that
$m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. It is showed in [15] that the minus
order and the Mitsch order coincide on regular modules.

Although the minus order on various algebraic structures has been an area of
intense research in the past few years, the main properties of the Mitsch
order is not studied in detail. Motivated by the aforementioned works on the
minus order, the goal of this paper is to investigate properties of the Mitsch
order on arbitrary modules. The paper is organized as follows: In Section 2,
we show that the Mitsch relation is a partial order on modules, and give some
characterizations of this partial order in various ways. We determine the way
how to define the Mitsch order from a module to its factor modules. In Section
3, the relations between the Mitsch order and the other well-known orders on
modules are studied and the following diagram is obtained:
Direct Sum OrderMinus Partial OrderJones OrderMitsch OrderSpace Pre-Order
We supply examples to show that all implications in this diagram are strict.
Section 4 deals with lattice properties of the Mitsch order on modules. The
maximal elements of modules with respect to the Mitsch order are determined.
Finally, we focus on the compatibility of the Mitsch order with the addition
and multiplication.

##  2\. Basic Properties of the Mitsch Order

In [12], Mitsch generalized Hartwig’s definition of the minus partial order to
arbitrary semigroups. Suppose $a,b$ are two elements of an arbitrary semigroup
$\mathcal{S}$. Then we write

$a\leq_{M}b\text{\quad if\quad}a=xb=by\text{ and }xa=a$ (4)

for some elements $x,y\in\mathcal{S}^{1}$, where $S^{1}$ denotes the set $S$
if $S$ has an identity. Mitsch proved that $\leq_{M}$ is indeed a partial
order for any semigroup $\mathcal{S}$ and that for $a,b\in\mathcal{S}$,

$a\leq_{M}b\text{\quad if and only if\quad}a=xb=by\text{ and }xa=ay=a$ (5)

for some elements $x,y\in\mathcal{S}^{1}$. One can easily generalize this
definition to a ring $R$ as follows.

######  Definition 2.1.

Let $R$ be a ring and $a,b\in R$. We write $a\leq_{M}b$ if there exist $x,y\in
R$ such that $a=xb=by$ and $a=xa$.

It can be easily seen that the Mitsch order is a partial order on rings.

######  Proposition 2.2.

Let $M$ be a module and $m_{1},m_{2}\in M$. Then the following statements are
equivalent.

  1. (1)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and
$m_{1}=fm_{1}$.

  2. (2)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and
$m_{1}=m_{1}a$.

  3. (3)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and
$m_{1}=fm_{1}=m_{1}a$.

###### Proof.

(1) $\Rightarrow$ (2) If there exist $f\in S$ and $a\in R$ such that
$m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$, then $m_{1}=fm_{2}a=m_{1}a$.
(2) $\Rightarrow$ (3) If there exist $f\in S$ and $a\in R$ such that
$m_{1}=m_{2}a=fm_{2}$ and $m_{1}=m_{1}a$, then $m_{1}=m_{1}a=fm_{2}a=fm_{1}$.
(3) $\Rightarrow$ (1) It is obvious. ∎

Following the above ideas, we now consider the relation $\leq_{M}$ on a module
$M$ which was defined in [15].

######  Definition 2.3.

Let $M$ be a module and $m_{1},m_{2}\in M$. We write $m_{1}\leq_{M}m_{2}$ if
one of the equivalent conditions in Proposition 2.2 is satisfied. We call the
relation $\leq_{M}$ the Mitsch order on $M$.

In the sequel of this section, we examine the main properties of the Mitsch
order on modules in detail.

######  Theorem 2.4.

Let $M$ be a module. The Mitsch order is a partial order on $M$.

###### Proof.

Reflexivity: Obvious.
Antisymmetry: Let $m_{1},m_{2}\in M$ with $m_{1}\leq_{M}m_{2}$ and
$m_{2}\leq_{M}m_{1}$. Then there exist $f,g\in S$ and $a,b\in R$ such that

$m_{1}=m_{2}a=fm_{2},~{}~{}m_{1}=fm_{1}$

and

$m_{2}=m_{1}b=gm_{1},~{}~{}m_{2}=gm_{2}.$

Hence $m_{1}=m_{2}a=gm_{2}a=gm_{1}=m_{2}$.
Transitivity: Let $m_{1},m_{2},m_{3}\in M$ with $m_{1}\leq_{M}m_{2}$ and
$m_{2}\leq_{M}m_{3}$. Then there exist $f,g\in S$ and $a,b\in R$ such that

$m_{1}=m_{2}a=fm_{2},~{}~{}m_{1}=fm_{1}$

and

$m_{2}=m_{3}b=gm_{3},~{}~{}m_{2}=gm_{2}.$

Hence $m_{1}=m_{3}(ba)=(fg)m_{3}$ and
$m_{1}=fm_{1}=fm_{2}a=fgm_{2}a=(fg)m_{1}$ and so $m_{1}\leq_{M}m_{3}$.
Therefore the Mitsch order is a partial order on $M$. ∎

In the next result, we show that there are many different ways to express the
Mitsch order.

######  Theorem 2.5.

Let $M$ be a module and $m_{1},m_{2}\in M$. The following are equivalent.

  1. (1)

$m_{1}\leq_{M}m_{2}$.

  2. (2)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a^{n}=f^{n}m_{2}$ and
$m_{1}=f^{n}m_{1}$, for any $n\in\mathbb{Z}^{+}$.

  3. (3)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a^{n}=f^{n}m_{2}$ and
$m_{1}=m_{1}a^{n}$, for any $n\in\mathbb{Z}^{+}$.

  4. (4)

There exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a^{n}=f^{n}m_{2}$ and
$m_{1}=f^{n}m_{1}=m_{1}a^{n}$, for any $n\in\mathbb{Z}^{+}$.

###### Proof.

(1) $\Leftrightarrow$ (2) Let $m_{1}\leq_{M}m_{2}$, then there exist $f\in S$
and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. We show that
$m_{1}=m_{2}a^{n}=f^{n}m_{2}$ and $m_{1}=f^{n}m_{1}$ for any
$n\in\mathbb{Z}^{+}$. If $n=1$ there is nothing to show. Assume $n>1$ then
$m_{1}=fm_{1}=ffm_{1}=\cdots=\underbrace{ff\cdots
f\,}_{\text{$n$-times}}m_{1}=f^{n}m_{1}$,
$m_{1}=fm_{1}=ffm_{2}=\cdots=\underbrace{ff\cdots
f\,}_{\text{$n$-times}}m_{2}=f^{n}m_{2}$ and
$m_{1}=m_{2}a=m_{2}aa=\cdots=m_{2}\underbrace{aa\cdots
a\,}_{\text{$n$-times}}=m_{2}a^{n}$. The converse is clear.
(1) $\Leftrightarrow$ (3) and (1) $\Leftrightarrow$ (4) can be proved similar
to the proof of (1) $\Leftrightarrow$ (2). ∎

The following corollary is a direct consequence of Theorem 2.5 which is a new
characterization of the Mitsch order on rings.

######  Corollary 2.6.

Let $R$ be a ring and $a,b\in R$. The following are equivalent.

  1. (1)

$a\leq_{M}b$.

  2. (2)

There exist $x,y\in R$ such that $a=bx^{n}=y^{n}b$ and $a=y^{n}a$, for any
$n\in\mathbb{Z}^{+}$.

  3. (3)

There exist $x,y\in R$ such that $a=bx^{n}=y^{n}b$ and $a=ax^{n}$, for any
$n\in\mathbb{Z}^{+}$.

  4. (4)

There exist $x,y\in R$ such that $a=bx^{n}=y^{n}b$ and $a=y^{n}a=ax^{n}$, for
any $n\in\mathbb{Z}^{+}$.

In the next result, we give some basic properties of the Mitsch order.

######  Proposition 2.7.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq_{M}m_{2}$, then the
following hold.

  1. (1)

$l_{S}(m_{2})\subseteq l_{S}(m_{1})$.

  2. (2)

$r_{R}(m_{2})\subseteq r_{R}(m_{1})$.

  3. (3)

$m_{1}R\subseteq m_{2}R$.

  4. (4)

$Sm_{1}\subseteq Sm_{2}$.

###### Proof.

It is straightforward. ∎

We now give an example to show that the converse of Proposition 2.7 need not
be true.

######  Example 2.8.

Let $R=M_{2}(\mathbb{Z})$, and $M=R_{R}$. Consider
$m_{1}=\begin{bmatrix}3&0\\\ 0&3\end{bmatrix}$,
$m_{2}=\begin{bmatrix}-1~{}&2\\\ 2&-1\end{bmatrix}$ and
$a=f=\begin{bmatrix}1&2\\\ 2&1\end{bmatrix}$ $\in M$. Since
$m_{1}=m_{2}a=fm_{2}$ we have $l_{S}(m_{2})\subseteq l_{S}(m_{1})$,
$r_{R}(m_{2})\subseteq r_{R}(m_{1})$, $m_{1}R\subseteq m_{2}R$,
$Sm_{1}\subseteq Sm_{2}$. On the other hand, it can be seen that
$m_{1}\nleq_{M}m_{2}$.

Let $M$ be a module and $N$ a submodule of $M$. It is natural to ask if it is
possible to extend the Mitsch order from $N$ to $M$ and vice versa. The
following result answers these questions.

######  Proposition 2.9.

Let $M$ be a module, $N$ a submodule of $M$ and $m_{1},m_{2}\in N$.

  1. (1)

If $N$ is a direct summand of $M$ and $m_{1}\leq_{M}m_{2}$ on $N$, then
$m_{1}\leq_{M}m_{2}$ on $M$.

  2. (2)

If $N$ is fully invariant in $M$ and $m_{1}\leq_{M}m_{2}$ on $M$, then
$m_{1}\leq_{M}m_{2}$ on $N$.

###### Proof.

(1) Let $m_{1}\leq_{M}m_{2}$ on $N$. Then there exist $f\in$ End${}_{R}(N)$
and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. If $N$ is a
direct summand of $M$, then there exists a submodule $K$ of $M$ such that
$M=N\oplus K$. Hence $f\oplus 0_{K}\in S$ and $m_{1}=m_{2}a=(f\oplus
0_{K})m_{2}$ and $m_{1}=(f\oplus 0_{K})m_{1}$. Therefore $m_{1}\leq_{M}m_{2}$
on $M$.
(2) Since $m_{1}\leq_{M}m_{2}$ on $M$, there exist $f\in S$ and $a\in R$ such
that $m_{1}=fm_{2}=m_{2}a$ and $m_{1}=fm_{1}$. The submodule $N$ being fully
invariant yields that $f{\mid_{N}}\in$ End${}_{R}(N)$. This completes the
proof. ∎

Let $M$ be a module and $N$ a submodule of $M$. In the next result, we
determine under what conditions the Mitsch order on $M$ can be described on
$M/N$.

######  Proposition 2.10.

Let $M$ be a module and $N$ a submodule of $M$. If $N$ is fully invariant,
then Mitsch order on $M$ can be defined on $M/N$.

###### Proof.

Let $M$ be a module, $N$ a fully invariant submodule of $M$ and
$m_{1},m_{2}\in M$. Assume that $m_{1}\leq_{M}m_{2}$ on $M$. Then there exists
$f\in S$ and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$.
Define $g\in$ End${}_{R}(M/N)$ as $g(m+N)=fm+N$ for $m+N\in M/N$. Hence $g$ is
well defined map on $M/N$. In fact, $m+N=m_{1}+N$ implies $m-m_{1}\in N$.
Since $N$ fully invariant, $f(m-m_{1})=f(m)-f(m_{1})\in f(N)\subseteq N$. Thus
$g(m+N)=f(m)+N=f(m_{1})+N$. Since $f$ is an endomorphism of $M$, $g$ is also
an endomorphism of $M/N$. Let $m_{1}+N$, $m_{2}+N\in M/N$ with
$m_{1}\leq_{M}m_{2}$ on $M$. By assumption $m_{1}=m_{2}a=fm_{2}$ and
$m_{1}=fm_{1}$. Then $m_{1}+N=m_{2}a+N=(m_{2}+N)a$ and
$m_{1}+N=fm_{2}+N=g(m_{2}+N)$. $m_{1}+N=fm_{1}+N=g(m_{1}+N)$. It entails that
$m_{1}+N\leq_{M}m_{2}+N$ on $M/N$. ∎

######  Corollary 2.11.

Let $R$ be a ring and $I$ an ideal of $R$. Then the Mitsch order on $R$ can be
defined on $R/I$.

##  3\. Orders Versus The Natural Partial Order On Modules

In this section, we investigate the relations between the Mitsch order and the
other well-known orders, namely, the minus order, the Jones order, the direct
sum order and the space pre-order on modules.

Let $M$ be a module and $m_{1},m_{2}\in M$. In [15], recall that
$m_{1}\leq^{-}m_{2}$ if there exists $\varphi\in M^{*}$ such that
$m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$, and $\varphi
m_{1}=\varphi m_{2}$. The relation $\leq^{-}$ is called the _minus order_ on
$M$.

######  Proposition 3.1.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq^{-}m_{2}$, then
$m_{1}\leq_{M}m_{2}$. The converse holds if $M$ is a regular module.

###### Proof.

If $m_{1}\leq^{-}m_{2}$, then there exists $\varphi\in M^{*}$ such that
$m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$, and $\varphi
m_{1}=\varphi m_{2}$. Take $a=\varphi m_{1}\in R$ and $f=m_{1}\varphi\in S$.
Then $m_{1}=m_{1}\varphi m_{1}=m_{2}\varphi m_{1}=m_{2}a$, $m_{1}=m_{1}\varphi
m_{1}=m_{1}\varphi m_{2}=fm_{2}$ and $m_{1}=m_{1}\varphi m_{1}=fm_{1}$.
Therefore $m_{1}\leq_{M}m_{2}$. Conversely, if $M$ is a regular module and
$m_{1}\leq_{M}m_{2}$, then from [15, Theorem 5.6] we have
$m_{1}\leq^{-}m_{2}$. ∎

The following example shows that the regularity condition in Proposition 3.1
is not superfluous.

######  Example 3.2.

Consider $M=\mathbb{Z}_{12}$ as a $\mathbb{Z}$-module. Since $M^{*}=0$, $M$ is
not a regular module. Let $a=3\in\mathbb{Z}$ and the homomorphism $f\in S$
defined by $f(\bar{n})=\bar{n}a$ for all $\bar{n}\in M$ we have
$\bar{6}=\bar{2}a=f(\bar{2})$ and $\bar{6}=f(\bar{6})$ which yields that
$\bar{6}\leq_{M}\bar{2}$. But since $M^{*}=0$, $\bar{6}\nleq^{-}\bar{2}$.

In [15], the Jones order on modules is defined in the following way: Let $M$
be a module and $m_{1},m_{2}\in M$. Then we write $m_{1}\leq_{J}m_{2}$ if
there exist $f^{2}=f\in S$ and $a^{2}=a\in R$ such that $m_{1}=m_{2}a=fm_{2}$.
In the next result, we obtain the relation between the Mitsch order and the
Jones order.

######  Proposition 3.3.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq_{J}m_{2}$, then
$m_{1}\leq_{M}m_{2}$.

###### Proof.

Asume that $m_{1}\leq_{J}m_{2}$. Then $m_{1}=m_{2}a=fm_{2}$ for some
$f^{2}=f\in S$ and $a^{2}=a\in R$. From $m_{1}=fm_{2}$ we obtain
$fm_{1}=fm_{2}=m_{1}$. Therefore $m_{1}\leq_{M}m_{2}$. ∎

The following example shows that there are Mitsch ordered elements which are
not Jones ordered in a module.

######  Example 3.4.

Let $C$ be the commutative semigroup on generators $a$, $b$ and $c$ with
relations $ac=a=cb$ as in [9, Example 2.9]. Consider the semigroup ring
$R=\mathbb{Z}_{2}[C]$. Being $ac=a=cb$ implies $ab=acb=a^{2}$. It can be
listed the distinct elements of $C$ and $R$ as

$C=\\{a,a^{2},\dots,a^{n}~{},\dots;b,b^{2},\dots,b^{n},\dots;c,c^{2},\dots,c^{n},\dots\\}$
$R=\\{\sum_{i}\sum_{j}\sum_{k}a^{i}b^{j}c^{k}\mid a,b,c\in C\\}.$

It is easy to see that $C$ is idempotent-free and the idempotents of $R$ are
trivial. The commutativity of $C$ and $R$ entails that in the equalities
$m_{1}=m_{2}d=fm_{2},$ $m_{1}=fm_{1}$ we replace $m_{1}$ by $a$, $m_{2}$ by
$b$, $d$ by $c$ and $f$ by $c$ to get $m_{1}=m_{2}d$ to $a=bc$, $m_{1}=fm_{2}$
to $a=cb$ and $m_{1}=fm_{1}$ to $a=ca$. Then $m_{1}=m_{2}d=fm_{2}$,
$m_{1}=fm_{1}$ is $a=bc=cb,$ $a=ca$. Hence we have $a\leq_{M}b$. Since $R$ has
only trivial idempotents $\\{0,1\\}$, $a\nleq_{J}b$.

A natural question is to ask under which conditions Mitsch ordered elements of
a module are Jones ordered.

######  Theorem 3.5.

Let $M$ be a module and $m_{1},m_{2}\in M$. Then we have the following.

  1. (1)

If $m_{1}\leq_{M}m_{2}$ and $r_{R}(m_{2})=l_{S}(m_{2})=0$, then
$m_{1}\leq_{J}m_{2}$.

  2. (2)

If $m_{1}\leq_{M}m_{2}$ and $m_{1}$ is regular, then $m_{1}\leq_{J}m_{2}$.

###### Proof.

(1) Let $m_{1}\leq_{M}m_{2}$, then there exist $f\in S$ and $a\in R$ such that
$m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}=m_{1}a$. Hence
$fm_{2}=fm_{1}=f^{2}m_{2}$ and $m_{2}a=m_{1}a=m_{1}a^{2}$. Thus
$(f-f^{2})m_{2}=0$ and $m_{2}(a-a^{2})=0$. Since $r_{R}(m_{2})=l_{S}(m_{2})=0$
then $f\in S$ and $a\in R$ are idempotent elements. Therefore
$m_{1}\leq_{J}m_{2}$.
(2) Let $m_{1}\in M$ be regular and $m_{1}\leq_{M}m_{2}$. The regularity of
$m_{1}$ yields there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi
m_{1}$. Then $m_{1}\varphi\in S$ is an idempotent and $\varphi m_{1}\in R$ is
an idempotent. Consider $m_{1}=m_{1}\varphi m_{1}=(m_{2}r)(\varphi
m_{1})=m_{2}(r\varphi m_{1})=m_{2}e$ where $e=r\varphi m_{1}$. We claim that
$e^{2}=e$. In fact $e^{2}=(r\varphi m_{1})(r\varphi
m_{1})=(r\varphi)(m_{1}r)(\varphi m_{1})=(r\varphi)(m_{1})(\varphi
m_{1})=r\varphi m_{1}=e$. Next we consider $m_{1}=m_{1}\varphi
m_{1}=(m_{1}\varphi)(sm_{2})=(m_{1}\varphi s)m_{2}=fm_{2}$ where $m_{1}\varphi
s=f\in S$ and we claim $f^{2}=f$. In fact $f^{2}=(m_{1}\varphi s)(m_{1}\varphi
s)=(m_{1}\varphi)(sm_{1})(\varphi s)=(m_{1}\varphi)(m_{1})(\varphi
s)=m_{1}\varphi s=f$. The net result is $m_{1}=m_{2}e$ and $e^{2}=e\in R$ and
$m_{1}=fm_{2}$ and $f^{2}=f\in S$. It follows that $m_{1}\leq_{J}m_{2}$,
$m_{1}=fm_{2}=m_{2}e=fm_{1}e=fm_{2}e$. ∎

Recall that the direct sum order on modules is defined in [15] as follows: Let
$M$ be a module and $m_{1},m_{2}\in M$. Then $m_{1}\leq^{\oplus}m_{2}$ if
$m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$.

######  Proposition 3.6.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq_{M}m_{2}$, then
$m_{1}\leq^{\oplus}m_{2}$. The converse holds if $M$ is a regular module.

###### Proof.

If $m_{1}\leq_{M}m_{2}$, then $m_{1}R\subseteq m_{2}R$ and
$(m_{2}-m_{1})R\subseteq m_{2}R$ from Proposition 2.7(3). It is clear that
$m_{2}R=m_{1}R+(m_{2}-m_{1})R$. Let $m\in m_{1}R\cap(m_{2}-m_{1})R$ then there
exists $a,b\in R$ such that $m=m_{1}a=(m_{2}-m_{1})b$. Since
$m_{1}\leq_{M}m_{2}$, $m_{1}=m_{1}r=m_{2}r$ for some $r\in R$. Thus
$m=m_{2}ra=(m_{2}-m_{2}r)b$. Hence $m_{2}(ra+rb-b)=0$. From Proposition 2.7(1)
$m_{1}(ra+rb-b)=0=m_{1}a+m_{1}b-m_{1}b=m_{1}a=m$. Therefore
$m_{1}R\cap(m_{2}-m_{1})R=0$ and $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$. The
converse is clear from [15, Theorem 5.11] and Proposition 3.1. ∎

As the following example shows, the condition being regular is not superfluous
in Proposition 3.6.

######  Example 3.7.

Let $F$ be a field and $R=\left\\{\begin{bmatrix}a&0&b\\\ 0&c&d\\\
0&0&e\end{bmatrix}\mid a,b,c,d,e\in F\right\\}$ be the subring of $U_{3}(F)$
the ring of $3\times 3$ upper triangular matrices over $F$.
Let $m_{1}=\begin{bmatrix}0&0&1\\\ 0&0&1\\\ 0&0&0\end{bmatrix}$ and
$m_{2}=\begin{bmatrix}0&0&1\\\ 0&1&1\\\ 0&0&0\end{bmatrix}$, and so
$m_{2}-m_{1}=\begin{bmatrix}0&0&0\\\ 0&1&0\\\ 0&0&0\end{bmatrix}$. Then
$m_{1}R=\left\\{\begin{bmatrix}0&0&a\\\ 0&0&a\\\ 0&0&0\end{bmatrix}\mid a\in
F\right\\}$, $m_{2}R=\left\\{\begin{bmatrix}0&0&b\\\ 0&c&d\\\
0&0&0\end{bmatrix}\mid b,c,d\in F\right\\}$ and
$(m_{2}-m_{1})R=\left\\{\begin{bmatrix}0&0&0\\\ 0&s&t\\\
0&0&0\end{bmatrix}\mid s,t\in F\right\\}$. Since $F$ is a field, it is easy to
check that $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$. So $m_{1}\leq^{\oplus}m_{2}$.
On the other hand, we claim that there is not an element $f$ in $R$ such that
$m_{1}=fm_{2}$. Otherwise, let $f=\begin{bmatrix}a&0&b\\\ 0&z&u\\\
0&0&e\end{bmatrix}\in R$ such that $\begin{bmatrix}0&0&1\\\ 0&0&1\\\
0&0&0\end{bmatrix}=\begin{bmatrix}a&0&b\\\ 0&z&u\\\
0&0&e\end{bmatrix}\begin{bmatrix}0&0&1\\\ 0&1&1\\\
0&0&0\end{bmatrix}=\begin{bmatrix}0&0&a\\\ 0&z&z\\\ 0&0&0\end{bmatrix}$. It
entails that $1=0$. This contradiction shows that $m_{1}\nleq_{M}m_{2}$.

In [11], the space pre-order on complex matrices was introduced by Mitra.
Recently, the space pre-order is extended to modules in [16]. Let $M$ be a
module and $m_{1},m_{2}\in M$. Then $m_{1}\leq_{S}m_{2}$ if $Sm_{1}\subseteq
Sm_{2}$ and $m_{1}R\subseteq m_{2}R$.

######  Proposition 3.8.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq_{M}m_{2}$, then
$m_{1}\leq_{S}m_{2}$.

###### Proof.

It is clear from Proposition 2.7. ∎

One may suspect that the converse of Proposition 3.8 holds. But the following
example erases the possibility.

######  Example 3.9.

Let $R=\begin{bmatrix}\mathbb{Z}&2\mathbb{Z}\\\
2\mathbb{Z}&\mathbb{Z}\end{bmatrix}$, and $M=R_{R}$. Consider
$m_{1}=\begin{bmatrix}9&6\\\ 6&3\end{bmatrix}$, $m_{2}=\begin{bmatrix}1&2\\\
2&1\end{bmatrix}$, $a=\begin{bmatrix}1&0\\\ 4&3\end{bmatrix}$ and
$f=\begin{bmatrix}1&4\\\ 0&3\end{bmatrix}$ $\in M$. Since
$m_{1}=m_{2}a=fm_{2}$ we have $m_{1}\leq_{S}m_{2}$. It can be seen that
$m_{1}\nleq_{M}m_{2}$.

We close this section by obtaining the relation between the minus order and
the Jones order. So the following result completes the diagram which is given
at the end of Introduction.

######  Proposition 3.10.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq^{-}m_{2}$, then
$m_{1}\leq_{J}m_{2}$.

###### Proof.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq^{-}m_{2}$, then
there exist $f^{2}=f\in S,a^{2}=a\in R$ such that
$m_{1}=fm_{1}=fm_{2}=m_{1}a=m_{2}a$, by [15, Lemma 5.2]. ∎

The next example shows that the converse of Proposition 3.10 need not be true.

######  Example 3.11.

Let $n$ be a positive integer and consider the $\mathbb{Z}$-module
$\mathbb{Z}_{n}$. Then the dual module $\mathbb{Z}_{n}^{*}=0$. Let $0\neq
m\in\mathbb{Z}_{n}$. On the one hand, since $\mathbb{Z}_{n}^{*}=0$, we have
$m\nleq^{-}m$. On the other hand, clearly, $m\leq_{J}m$.

##  4\. Lattice Properties of the Mitsch Order

In this section, we study lattice properties of the Mitsch order on modules.
Hence we present various characterizations of the Mitsch order as an
application. We characterize the maximal elements of modules with respect to
the Mitsch order. We investigate the compatibility of the Mitsch order with
the addition and multiplication. We give examples to show that the Mitsch
order is not compatible with the addition and multiplication in general.

Let $R$ and $S$ be rings, ${}_{R}V_{S}$ and ${}_{S}W_{R}$ be bimodules. Then
the 4-tuple $(R,V,W,S)$ is said to be a _Morita context_ if there exist
products $V\times W\rightarrow R$ defined by $(v,w)\mapsto vw$ and $W\times
V\rightarrow S$ defined by $(w,v)\mapsto wv$. With these requirements,
$\begin{bmatrix}R&V\\\ W&S\end{bmatrix}$ is an associative ring using matrix
operations and it is called a _Morita context ring_ [13].

######  Remark 4.1.

Let $M$ be a module. Then ${}_{S}M_{R}$ is a bimodule. Also, $M^{*}$ is a
right $S$-module via composition of maps, and $M^{*}$ becomes a left
$R$-module as follows: For $\varphi\in M^{*}$ and $r\in R$ define $r\varphi\in
M^{*}$ by $(r\varphi)x=r(\varphi x)$ for all $x\in M$. Hence we have
${}_{S}M_{R}$ and ${}_{R}(M^{*})_{S}$ bimodules. Also, if $m\in M$ and
$\varphi\in M^{*}$, then $\varphi m\in R$ and dually $m\varphi\in S$ with
$(m\varphi)x=m(\varphi x)\equiv m\varphi x$ for all $x\in M$. If $m\in M$ is
regular, then $m=m\varphi m$ for some $\varphi\in M^{*}$, so we have
idempotent elements $\varphi m\in R$ and $m\varphi\in S$. It is easy to verify
that $(R,M^{*},M,S)$ is a Morita context, called the standard context of the
module $M$.

In [2], Chen et al. defined left invertible and right invertible elements of a
module by using the Morita context as follows. Let $(R,M^{*},M,S)$ be a
standard context of the module $M$. Then

$m\in M~{}~{}\text{is
called}~{}~{}\begin{cases}{\textit{l}eft~{}invertible}~{}\text{if}~{}\varphi
m=1_{R}~{}\text{for some}~{}\varphi\in M^{*},\\\
{\textit{r}ight~{}invertible}~{}\text{if}~{}m\varphi=1_{S}~{}\text{for
some}~{}\varphi\in M^{*}.\end{cases}$

######  Theorem 4.2.

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq_{M}m_{2}$ then the
following hold.

  1. (1)

If $m_{1}$ is left invertible, then $m_{1}=m_{2}$.

  2. (2)

If $m_{1}$ is right invertible, then $m_{1}=m_{2}$.

  3. (3)

If $m_{2}\in m_{1}R$, then $m_{1}=m_{2}$.

  4. (4)

If $m_{2}\in Sm_{1}$, then $m_{1}=m_{2}$.

  5. (5)

If $m_{2}\in N$ for a submodule $N$ of $M$, then $m_{1}\in N$.

  6. (6)

$m_{2}-m_{1}\leq_{M}m_{2}$.

###### Proof.

(1) Let $m_{1}\leq_{M}m_{2}$ and $m_{1}$ is left invertible. Then there exist
$a\in R$ and $f\in M^{*}$ such that $m_{1}=m_{1}a=m_{2}a$ and $fm_{1}=1_{R}$.
Hence $a=fm_{1}a=fm_{1}=1_{R}$. Therefore $m_{1}=m_{2}$.
(2) It can be proved similar to the proof of (1).
(3) Let $m_{1}\leq_{M}m_{2}$ and $m_{2}\in m_{1}R$. Then there exist $b\in R$
and $f\in S$ such that $m_{2}=m_{1}b$ and $fm_{1}=fm_{2}=m_{1}$. Thus
$m_{2}=m_{1}b=fm_{1}b=fm_{2}=m_{1}$.
(4) It can be proved similar to the proof of (3).
(5) It is clear.
(6) Let $m_{1}\leq_{M}m_{2}$. Then $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$
for some $f\in S$ and $a\in R$. Therefore $m_{2}-m_{1}=m_{2}(1-a)=(1-f)m_{2}$
and $m_{2}-m_{1}=(1-f)(m_{2}-m_{1})$. Hence $m_{2}-m_{1}\leq_{M}m_{2}$. ∎

Note that zero is the minimum element of the poset $(M,\leq_{M})$. In the
following, we determine the maximal elements of the module with respect to the
Mitsch order.

######  Theorem 4.3.

Let $M$ be a module and $m\in M$. Then $m$ is a maximal element of the the
poset $(M,\leq_{M})$ if any of the following holds.

  1. (1)

m is left invertible.

  2. (2)

m is right invertible.

  3. (3)

For every $b\in r_{R}(m)$, $b+1$ is invertible.

  4. (4)

For every $g\in l_{S}(m)$, $g+1$ is invertible.

  5. (5)

$r_{R}(m)=0$.

  6. (6)

$l_{S}(m)=0$.

###### Proof.

(1) and (2) follow from Theorem 4.2.
(3) Let $n\in M$ and $m\leq_{M}n$, then there exists $a\in R$ such that
$m=na=ma$. Hence $m(a-1)=0$. Therefore $a\in R$ is invertible and $na=ma$
implies that $n=m$. Thus m is a maximal element of the poset $(M,\leq_{M})$.
(4) It can be proved similar to the proof of (3).
(5) and (6) are clear from (3) and (4), respectively. ∎

The following corollary is a direct consequence of Theorem 4.2 and Theorem
4.3.

######  Corollary 4.4.

Let $M$ be module.

  1. (1)

If $M_{R}$ or ${}_{S}M$ is a faithful module, then every nonzero element of
$M$ is maximal with respect to $\leq_{M}$.

  2. (2)

If $M_{R}$ or ${}_{S}M$ is a cyclic module generated by m, then $m$ is a
maximal element of $M$ with respect to $\leq_{M}$.

  3. (3)

All torsion-free elements of $M$ are maximal with respect to $\leq_{M}$.

One may suspect that a module $M$ becomes a lattice with respect to the Mitsch
order. But the following remark reveals that this is not the case.

######  Remark 4.5.

Let $M$ be a regular module and $m_{1},m_{2}\in M$. By Proposition 3.1,
$m_{1}\leq_{M}m_{2}$ implies $m_{1}\leq^{-}m_{2}$. Assume that $m_{1}$ and
$m_{2}$ have an upper bound $m\in M$ in the lattice of the Mitsch order on
$M$. Then $m$ is also an upper bound of $m_{1}$ and $m_{2}$ in the lattice of
minus order on $M$. It is known by [1, Example 3.6] that the minus order on a
regular module $M$ need not form a lattice on $M$. Therefore $M$ need not be a
lattice with respect to the Mitsch order in general.

In the following example, we show that the Mitsch order is not always
compatible with the scalar multiplication.

######  Example 4.6.

Let $R=\begin{bmatrix}\mathbb{Z}&\mathbb{Z}_{6}\\\
0&\mathbb{Z}_{6}\end{bmatrix}$ and $M=R_{R}$. Consider
$m_{1}=\begin{bmatrix}2&\bar{2}\\\ 0&\bar{3}\end{bmatrix}$ and
$m_{2}=\begin{bmatrix}2&\bar{0}\\\ 0&\bar{1}\end{bmatrix}$ in $M$. Then for
$a=\begin{bmatrix}1&\bar{1}\\\ 0&\bar{3}\end{bmatrix}$,
$f=\begin{bmatrix}1&\bar{2}\\\ 0&\bar{3}\end{bmatrix}\in R$ we have
$m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. Hence $m_{1}\leq_{M}m_{2}$. Set
$g=\begin{bmatrix}2&\bar{0}\\\ 0&\bar{3}\end{bmatrix}\in R$. Since there is no
solution for the equation $\overline{3x}=4$ in $\mathbb{Z}_{6}$, we obtain
$gm_{1}\nleq_{M}gm_{2}$.

We have the following result for which the Mitsch order is compatible with the
scalar multiplication. Recall that a ring $R$ is called a right (resp., left)
duo ring if every right (resp., left) ideal of $R$ is an ideal; equivalently
$Ra\subseteq aR~{}(\mbox{resp.,}~{}aR\subseteq Ra)$ for every $a\in R$.

######  Proposition 4.7.

Let $M$ be a module and $m_{1},m_{2}\in M$. Then we have the following.

  1. (1)

$m_{1}\leq_{M}m_{2}$ if and only if $gm_{1}\leq_{M}gm_{2}$ for every
invertible element $g\in S$.

  2. (2)

$m_{1}\leq_{M}m_{2}$ if and only if $m_{1}b\leq_{M}m_{2}b$ for every
invertible element $b\in R$.

  3. (3)

If $m_{1}\leq_{M}m_{2}$ and $b\in C(R)$, then $m_{1}b\leq_{M}m_{2}b$.

  4. (4)

If $m_{1}\leq_{M}m_{2}$ and $g\in C(S)$, then $gm_{1}\leq_{M}gm_{2}$.

  5. (5)

If $m_{1}\leq_{M}m_{2}$ and $R$ is a right duo ring, then
$m_{1}b\leq_{M}m_{2}b$ for all $b\in R$.

  6. (6)

If $m_{1}\leq_{M}m_{2}$ and $S$ is a left duo ring, then
$gm_{1}\leq_{M}gm_{2}$ for all $g\in S$.

###### Proof.

(1) Assume that $m_{1}\leq_{M}m_{2}$. Then there exist $f\in S$ and $a\in R$
such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. Let $g\in S$ be
invertible. Then $gm_{1}=gm_{2}a=gfm_{2}$ and $gm_{1}=gfm_{1}$. Since $g$ is
invertible, $gm_{1}=gm_{2}a=gfg^{-1}gm_{2}$. Therefore $gm_{1}\leq_{M}gm_{2}$.
Conversely, if $gm_{1}\leq_{M}gm_{2}$ for an invertible $g\in S$, then
$g^{-1}gm_{1}\leq_{M}g^{-1}gm_{2}$ and so $m_{1}\leq_{M}m_{2}$.
(2) It can be proved similar to the proof of (1).
(3) Assume that $m_{1}\leq_{M}m_{2}$. Then there exist $f\in S$ and $a\in R$
such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$. Since $b\in C(R)$,
$m_{1}b=m_{2}ab=m_{2}ba=fm_{2}b$ and $m_{1}b=fm_{1}b$. Therefore
$m_{1}b\leq_{M}m_{2}b$.
(4) It can be proved similar to the proof of (3).
(5) Let $m_{1}\leq_{M}m_{2}$ and $b\in R$. Then $m_{1}=fm_{2}=m_{2}a$ and
$m_{1}=m_{1}a$ for some $f\in S$ and $a\in R$. It follows that
$m_{1}b=fm_{2}b=m_{2}ab=m_{1}ab$. Since $R$ is right duo, there exists $c\in
R$ such that $ab=bc$. Thus $m_{1}b\leq_{M}m_{2}b$.
(6) It can be proved similar to the proof of (5). ∎

In Example 4.6, we show that the Mitsch order is not compatible with the
scalar multiplication. Following example shows that the Mitsch order also is
not compatible with addition.

######  Example 4.8.

Let $R=M_{2}(\mathbb{Q})$ and $M=R_{R}$. Consider $m_{1}=\begin{bmatrix}2&5\\\
0&0\end{bmatrix}$, $m_{2}=\begin{bmatrix}1&5\\\ 1&0\end{bmatrix}$,
$f=\begin{bmatrix}1&1\\\ 0&0\end{bmatrix}$ and $a=\begin{bmatrix}0&0\\\
\frac{2}{5}&1\end{bmatrix}$ $\in M$. Since $m_{1}=m_{2}a=fm_{2}$ and
$m_{1}=fm_{1}$ we have $m_{1}\leq_{M}m_{2}$ and $0\leq_{M}m_{2}$. Since there
is no $b\in R$ such that $(m_{2}+m_{2})b=m_{1}b=m_{1}$ we have
$m_{1}\nleq_{M}m_{2}+m_{2}$.

We give the following result in which cases the Mitsch order is compatible
with the addition.

######  Proposition 4.9.

Let $M$ be a module and $m_{1},m_{2}\in M$. Then the following statements are
equivalent.

  1. (1)

$m_{1}\leq_{M}m_{2}$.

  2. (2)

$nm_{1}\leq_{M}m_{2}$ for all $n\in\mathbb{Z}$.

  3. (3)

$nm_{1}\leq_{M}nm_{2}$ for all $n\in\mathbb{Z}$.

###### Proof.

It is clear. ∎

######  Proposition 4.10.

Let $M$ be a module and $m\in M$.

  1. (1)

If $mR$ is a fully invariant submodule of $M$, then $fm\leq_{M}m$ for every
$f^{2}=f\in S$.

  2. (2)

If $mR$ is a homomorphic image of $M$, then $me\leq_{M}m$ for every
$e^{2}=e\in R$.

###### Proof.

(1) Assume that $f^{2}=f\in S$ and $mR$ is fully invariant in $M$. Then
$fm=ffm$ and since $fm\in mR$, there exists $a\in R$ such that $fm=ma$.
Therefore $fm\leq_{M}m$.
(2) Similar to the proof of (1). ∎

The next result is a direct consequence of Proposition 4.10.

######  Corollary 4.11.

Let $M$ be a module.

  1. (1)

If $M$ is duo, then $fm\leq_{M}m$ for all $m\in M$ and $f^{2}=f\in S$.

  2. (2)

If $M$ is 1-epi-retractable, then $me\leq_{M}m$ for all $m\in M$ and
$e^{2}=e\in R$.

## References

  * [1] S. R. Argun, T. P. Calci, S. Halicioglu, A. Harmanci and B. Ungor, Lattice properties of the minus order on regular modules, Rocky Mountain J. Math., 52(1)(2022), 15–27.
  * [2] H. Chen, W. K. Nicholson and Y. Zhou, Unit-regular modules, Glasg. Math. J., 60(1)(2018), 1-15.
  * [3] D. S. Djordjević, D. S. Rakić and J. Marovt, Minus partial order in Rickart rings, Publ. Math. Debrecen, 87(3-4)(2015), 291-305.
  * [4] G. Dolinar, B. Kuzma and J. Marovt A note on partial orders of Hartwig, Mitsch, and Šemrl, Appl. Math. Comput., 270 (2015), 711–713.
  * [5] M. P. Drazin, Generalizations of Fitting’s lemma in arbitrary associative rings, Comm. Algebra, 29(8)(2001), 3647-3675.
  * [6] R. E. Hartwig, How to partially order regular elements, Math. Japon., 25(1)(1980), 1-13.
  * [7] R. E. Hartwig and G. P. H. Styan, On some characterizations of the “star” partial ordering for matrices and rank substractivity, Linear Algebra Appl., 82(1986), 145-161.
  * [8] A. Hattori, A foundation of torsion theory for modules over general rings, Nagoya Math. J., 17(1960), 147-158.
  * [9] P. M. Higgins, The Mitsch order on a semigroup, Semigroup Forum, 49(2)(1994), 261-266.
  * [10] J. Marovt, On partial orders in Rickart rings, Linear Multilinear Algebra, 63(9)(2015), 1707-1723.
  * [11] S. K. Mitra, Matrix partial order through generalized inverses: unified theory, Linear Algebra Appl., 148(1991), 237-263.
  * [12] H. Mitsch, A natural partial order for semigroups, Proc. Amer. Math. Soc., 97(3)(1986), 384-388.
  * [13] K. Morita, Duality for modules and its applications to the theory of rings with minimum condition, Sci. Rep. Tokyo Kyoiku Daigaku Sect. A, 6(1958), 83-142.
  * [14] B. Ungor, S. Halicioglu, A. Harmanci and J. Marovt, Partial orders on the power sets of Baer rings, J. Algebra Appl., 19(2020), 2050011 (14 pp).
  * [15] B. Ungor, S. Halicioglu, A. Harmanci and J. Marovt, Minus partial order in regular modules, Comm. Algebra, 48(10)(2020), 4542-4553.
  * [16] B. Ungor, S. Halicioglu, A. Harmanci and J. Marovt, On properties of the minus partial order in regular modules, Publ. Math. Debrecen, 96(1-2)(2020), 149-159.
  * [17] J. Zelmanowitz, Regular modules, Trans. Amer. Math. Soc., 163(1972), 341-355.