Minus Partial Order in Regular Modules

TL;DR

This paper extends the minus partial order concept to regular modules, providing new characterizations and generalizations, thereby broadening its applicability beyond matrices and operators.

Contribution

It introduces the minus partial order in the context of regular modules and proves it forms a partial order, with multiple characterizations and generalizations.

Findings

Minus partial order is a partial order in regular modules.

Provides new characterizations of the minus partial order.

Generalizes known results to the module setting.

Abstract

The minus partial order is already known for sets of matrices over a field and bounded linear operators on arbitrary Hilbert spaces. Recently, this partial order has been studied on Rickart rings. In this paper, we extend the concept of the minus relation to the module theoretic setting and prove that this relation is a partial order when the module is regular. Moreover, various characterizations of the minus partial order in regular modules are presented and some well-known results are also generalized.

Equations

$$a{\le }^{-}b\text{if}{a}^{-}a=a^{-}b\text{and}a{a}^{-}=b{a}^{-}$$

$$l_R(a)=R(1_R-p),r_R(a)=(1_R-q)R,pa=pb,\text{and}aq=bq.$$

$$m=m\phi (m)\equiv m\phi m$$

$$a\phi a=a\phi (a)=aba=a$$

$$a=a\phi (a)=a\phi (1_{R}a)=a\phi (1_R)a=a{a}^{-}a.$$

$$(m\phi )(x)=m\phi (x)\equiv m\phi x,x\in M$$

$$m_1=m_1\phi m_1,m_1\phi =m_2\phi ,\phi m_1=\phi m_2$$

$$m_2=m_2\beta m_2,m_2\beta =m_1\beta ,\beta m_2=\beta m_1.$$

$$m_1=f_1{m}_1=f_1{m}_2=m_1{a}_1=m_2{a}_1.$$

$$f{m}_3=m_1+f_1{m}_3-f_1{f}_2{f}_1{m}_3=m_1+f_1(1_S-f_2)f_1{m}_3.$$

$$m_{3}a=m_1+m_3{a}_1(1_R-a_2)a_1.$$

$$l_R(a)=R(1_R-p),r_R(a)=(1_R-q)R,pa=pb,\text{and }aq=bq.$$

$$R(1_R-p)=l_R(p)\text{and}(1_R-p)R=r_R(p).$$

$$A\ast \le B\text{if}{A}^{\ast }A=A^{\ast }B\text{ and }\mathrm{Im}A\subseteq \mathrm{Im}B.$$

$$m_2=f_2{m}_1=f_2{m}_2{a}_1=m_2{a}_1=m_1.$$

$$m_1=f_1{m}_1=f_1{m}_2=m_1{a}_1=m_2{a}_1.$$

$$m_{1}a=m_1{a}_1=m_2{a}_1=m_2{a}_2{a}_1=m_3{a}_2{a}_1=m_3{a}_1=m_{3}a.$$

$$f{m}_1=f{m}_3.$$

$$f_1=m_1\phi \in S\text{and}{a}_1=\phi m_1\in R.$$

$$a{\le }_{J}b\text{if}a=pb=bq$$

$$a{\le }_{M}b\text{if}a=xb=by\text{ and }xa=a$$

$$a{\le }_{M}b\text{if and only if}a=xb=by\text{ and }xa=ay=a$$

$$m_1=m_{2}a=f{m}_2\text{and}{m}_1=f{m}_1=m_{1}a.$$

$$0=x{m}_1\phi f{m}_1=x{m}_1\phi m_1=x{m}_1.$$

$$0=m_{2}a\phi m_{1}x=m_1\phi m_{1}x=m_{1}x.$$

$$g{m}_1=g(m_{2}b)=(g{m}_2)b=m_{1}b=m_1$$

$$a{\le }^{\oplus }b\text{if}bR=aR\oplus (b-a)R$$

$$m_1{\le }^{\oplus }{m}_{2}if{m}_{2}R=m_{1}R\oplus (m_2-m_1)R.$$

$$x=m_{2}r+m_{1}r-m_{1}r=m_{1}r+(m_2-m_1)r\in m_{1}R+(m_2-m_1)R.$$

$$m_{1}R+(m_2-m_1)R\subseteq m_{2}R.$$

Full text

1. Introduction

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 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 [7] introduced the minus partial order $\leq^{-}$ on regular semigroups using generalized inverses. For a regular semigroup $\mathcal{S}$ and $a,b\in\mathcal{S}$, we write

[TABLE]

for some inner generalized inverse $a^{-}$ of $a$.

Let $B(\mathcal{H})$ be the algebra of all bounded linear operators on a Hilbert space $\mathcal{H}$. For an operator $A\in B(\mathcal{H})$, the symbols $\mathrm{Ker}$ $A$ and $\mathrm{Im}$ $A$ will denote the kernel and the image of $A$, respectively. It is known that $A\in B(\mathcal{H})$ is regular if and only if $\operatorname{Im}\,A=\overline{\operatorname{Im}\,A}$, i.e. the image of $A$ is closed (see for example [12]). Šemrl studied in [14] the minus partial order on $B(\mathcal{H})$. He did not want to restrict himself only to operators on $B(\mathcal{H})$ with closed images so he defined a new order $\leq_{S}$ on $B(\mathcal{H})$ in the following way: For $A,B\in B(\mathcal{H})$ we write $A\leq_{S}B$ if there exist idempotent operators $P,Q\in B(\mathcal{H})$ such that $\operatorname{Im}\,P=\overline{\operatorname{Im}\,A}$, $\mathrm{Ker}\,A=\mathrm{Ker}\,Q$, $PA=PB$, and $AQ=BQ$. Šemrl called this order the minus partial order on $B(\mathcal{H})$ and proved that this is indeed a partial order on $B(\mathcal{H})$ for a general Hilbert space $\mathcal{H}$. He also showed that the partial order $\leq_{S}$ is the same as Hartwig’s minus partial order $\leq^{-}$ when $\mathcal{H}$ is finite dimensional.

For a subset $\mathcal{A}$ of a ring $R$, $l_{R}(\mathcal{A})$ and $r_{R}(\mathcal{A})$ will denote the left annihilator and the right annihilator of $\mathcal{A}$ in $R$, respectively. If the subset $\mathcal{A}$ is a singleton, say $\mathcal{A}=\{a\}$, then again we simply write $l_{R}(a)$ and $r_{R}(a)$, respectively. A ring $R$ is called a Rickart ring if for every $a\in R$ there exist idempotent elements $p,q\in R$ such that $r_{R}(a)=p\cdot R$ and $l_{R}(a)=R\cdot q$. Note that every Rickart ring $R$ has the (multiplicative) identity and that the class of Rickart rings includes von Neumann algebras and rings with no proper zero divisors (see [2] or [9]).

Following Šemrl’s approach, the authors further generalized in [4] the minus partial order to Rickart rings. A new relation was introduced on a ring with identity: Let $R$ be a ring with the identity $1_{R}$ and $a,b\in R$. Then we write $a\leq^{-}b$ if there exist idempotent elements $p,q\in R$ such that

[TABLE]

It was proved that this relation $\leq^{-}$ is indeed a partial order when $R$ is a Rickart ring and that definitions (1) and (2) are equivalent when $R$ is a ring in which every element is regular, i.e. $R$ is a von Neumann regular ring.

In this paper, we will study the minus partial order in a more general setting. The goal of this paper is to extend the notion of the minus partial order to modules using their endomorphism rings. We show that this minus relation is a partial order when the module is regular. We also present various characterizations of the minus partial order in regular modules and generalize some well-known results.

Throughout this paper $R$ denotes an associative ring with identity $1_{R}$ and modules are unitary right $R$-modules. For a right $R$-module $M_{R}=M$, $S=$ End${}_{R}(M)$ is the ring of all right $R$-module endomorphisms of $M$. It is well known that $M$ is a left $S$ and 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.

2. Minus partial order in modules

Let $M$ be a right $R$-module with $S=$ End${}_{R}(M)$. For the sake of brevity, in the sequel, $S$ will stand for the endomorphism ring of the module $M$ considered. We will denote the identity map in $S$ by $1_{S}$. An element $m\in M$ is called (Zelmanowitz) regular if

[TABLE]

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) so that there exists $b\in R$ such that $a=aba$. Define the map $\varphi:R\rightarrow R$ with $\varphi(r)=br,r\in R$. Then $\varphi\in R^{*}=$ End${}_{R}(R)$. We have

[TABLE]

which yields that $a$ is regular in $R_{R}$ (in the sense of Zelmanowitz, see also [15]).

Let now $R$ be a ring with identity and suppose $a$ is a regular element in $R_{R}$ (in the sense of Zelmanowitz). Then there exists $\varphi\in R^{*}$ such that $a=a\varphi a$. Define $a^{-}=\varphi(1_{R})\in R$. Then

[TABLE]

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.

Remark 2.1**.**

[13*, Lemma B.47]** *Let $M$ be a module and $m\in M$ regular, say $m=m\varphi m$ where $\varphi\in M^{*}$. Then $e=\varphi m\in R$ is an idempotent, $mR\cong eR$ is projective, and $M=mR\oplus N$ where $N=\{n\in M\mid m\varphi n=0\}$.

Remark 2.2**.**

Let $M$ be a module. It is known that Hom${}_{R}(R,M)\cong M$. Let $m\in M$ be regular, say $m=m\varphi m$ where $\varphi\in M^{*}$. Then for the map $m\varphi:M\rightarrow M$ defined by

[TABLE]

we may conclude that $m\varphi\in S$ and that $m\varphi$ is an idempotent in $S$.

With the following definition we will introduce the concept of minus order in the setting of modules.

Definition 2.3**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. We write $m_{1}\leq^{-}m_{2}$ if there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$, and $\varphi m_{1}=\varphi m_{2}$. We call the relation $\leq^{-}$ the minus order on $M$.**

We will prove that when the module $M$ is regular, the relation $\leq^{-}$ is a partial order. First, let us present an auxiliary result and a new characterization of the minus order.

Proposition 2.4**.**

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

Proof.

Assume that $m_{1}\leq^{-}m_{2}$. Then there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$, and $\varphi m_{1}=\varphi m_{2}$. Hence $m_{1}=m_{1}\varphi m_{1}=m_{2}\varphi m_{1}=m_{2}\varphi m_{2}\in m_{2}R$ and thus $m_{1}R\subseteq m_{2}R$. ∎

Theorem 2.5**.**

Let $M$ be a module and $m_{1},m_{2}\in M$ with $m_{1}$ regular. Then $m_{1}\leq^{-}m_{2}$ if and only if there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$.

Proof.

Let $m_{1}\leq^{-}m_{2}$. Then there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$, $m_{1}\varphi=m_{2}\varphi$, and $\varphi m_{1}=\varphi m_{2}$. Let $f=m_{1}\varphi$ and $a=\varphi m_{1}$. Then $f^{2}=f\in S$ and $a^{2}=a\in R$. Clearly, $l_{S}(m_{1})\subseteq l_{S}(f)$. From $m_{1}=m_{1}\varphi m_{1}$ we obtain $(1_{S}-m_{1}\varphi)m_{1}=0$. So, $l_{S}(f)=S(1_{S}-m_{1}\varphi)\subseteq l_{S}(m_{1})$. Hence $l_{S}(m_{1})=l_{S}(f)$. Obviously, $r_{R}(m_{1})\subseteq r_{R}(a)$. Similarly, since $m_{1}=m_{1}\varphi m_{1}$, $m_{1}(1_{R}-\varphi m_{1})=0$, it follows $r_{R}(a)=(1_{R}-\varphi m_{1})R\subseteq r_{R}(m_{1})$. Thus $r_{R}(m_{1})=r_{R}(a)$. Also, $fm_{1}=m_{1}\varphi m_{1}=m_{1}\varphi m_{2}=fm_{2}$ and $m_{1}a=m_{1}\varphi m_{1}=m_{2}\varphi m_{1}=m_{2}a$.

Conversely, assume that there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$. Also, $m_{1}=m_{1}\varphi m_{1}$ for some $\varphi\in M^{*}$. Then $m_{1}=fm_{1}=m_{1}a$. Let $\beta=a\varphi f\in M^{*}$. Hence $m_{1}\beta m_{1}=m_{1}a\varphi fm_{1}=m_{1}\varphi m_{1}=m_{1}$. Also, we have $m_{1}\beta=m_{1}a\varphi f=m_{2}a\varphi f=m_{2}\beta$ and $\beta m_{1}=a\varphi fm_{1}=a\varphi fm_{2}=\beta m_{2}$. Therefore $m_{1}\leq^{-}m_{2}$. ∎

We are now in position to prove that the minus order $\leq^{-}$ is indeed a partial order when the module $M$ is regular.

Theorem 2.6**.**

Let $M$ be a regular module. Then the relation $\leq^{-}$ is a partial order on $M$.

Proof.

Reflexivity: Obvious.

Antisymmetry: Let $m_{1},m_{2}\in M$ with $m_{1}\leq^{-}m_{2}$ and $m_{2}\leq^{-}m_{1}$. Then there exist $\varphi,\beta\in M^{*}$ such that

[TABLE]

and

[TABLE]

By Proposition 2.4, $m_{1}R=m_{2}R$. Remark 2.1 yields that $M=m_{1}R\oplus N_{1}$ where $N_{1}=\{n\in M\mid m_{1}\varphi n=0\}$. Since $m_{1}\varphi(m_{1}-m_{2})=0$, $m_{1}-m_{2}\in N_{1}$, say $m_{1}-m_{2}=n\in N_{1}$. Hence $m_{1}=n+m_{2}$, and also $m_{2}=m_{1}r$ for some $r\in R$. Thus $m_{1}=n+m_{1}r$, so $n=m_{1}-m_{1}r\in N_{1}\cap m_{1}R=\{0\}$. This implies that $m_{1}=m_{1}r$. Therefore $m_{1}=m_{2}$.

Transitivity: Let $m_{1},m_{2},m_{3}\in M$ with $m_{1}\leq^{-}m_{2}$ and $m_{2}\leq^{-}m_{3}$. By Theorem 2.5, there exist $f_{1}^{2}=f_{1}\in S$, $a_{1}^{2}=a_{1}\in R$ such that $l_{S}(m_{1})=l_{S}(f_{1})$, $r_{R}(m_{1})=r_{R}(a_{1})$, $f_{1}m_{1}=f_{1}m_{2}$, $m_{1}a_{1}=m_{2}a_{1}$; and $f_{2}^{2}=f_{2}\in S$, $a_{2}^{2}=a_{2}\in R$ such that $l_{S}(m_{2})=l_{S}(f_{2})$, $r_{R}(m_{2})=r_{R}(a_{2})$, $f_{2}m_{2}=f_{2}m_{3}$, $m_{2}a_{2}=m_{3}a_{2}$. From $(1_{S}-f_{1})m_{1}=0$ and $m_{1}(1_{R}-a_{1})=0$ it follows

[TABLE]

Let $f=f_{1}+f_{1}f_{2}(1_{S}-f_{1})$ and $a=a_{1}+(1_{R}-a_{1})a_{2}a_{1}$. Then $f^{2}=f\in S$ and $a^{2}=a\in R$. We claim that

$l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{3}$, and $m_{1}a=m_{3}a$.

1. (i)

Clearly, $l_{S}(m_{1})=l_{S}(f_{1})\subseteq l_{S}(f)$. Let $g\in l_{S}(f)$. Then $gff_{1}=0$. It follows $gf_{1}=0$ and so $g\in l_{S}(f_{1})=l_{S}(m_{1})$. Hence $l_{S}(m_{1})=l_{S}(f)$. 2. (ii)

It is obvious that $r_{R}(m_{1})=r_{R}(a_{1})\subseteq r_{R}(a)$. Let $x\in r_{R}(a)$. Then $a_{1}ax=0$. This yields $a_{1}x=0$, i.e. $x\in r_{R}(a_{1})=r_{R}(m_{1})$, and thus $r_{R}(m_{1})=r_{R}(a)$. 3. (iii)

We will now prove $fm_{1}=fm_{3}$. By (3), $fm_{1}=m_{1}$. Similarly, $fm_{2}=m_{2}$. From the definition of $f$ we obtain, $fm_{3}=f_{1}m_{3}+f_{1}f_{2}m_{3}-f_{1}f_{2}f_{1}m_{3}$. Note that $f_{1}f_{2}m_{3}=f_{1}f_{2}m_{2}=f_{1}m_{2}=f_{1}m_{1}=m_{1}$. Then

[TABLE]

We claim that $f_{1}(1_{S}-f_{2})f_{1}=0$, that is, $f_{1}(1-f_{2})\in l_{S}(f_{1})=l_{S}(m_{1})$. By $f_{1}f_{2}m_{1}=f_{1}f_{2}m_{2}a_{1}=f_{1}m_{2}a_{1}=f_{1}m_{1}$, we obtain $f_{1}(1_{S}-f_{2})m_{1}=0$ and hence $f_{1}(1_{S}-f_{2})f_{1}=0$, as claimed. Thus $fm_{3}=m_{1}$ and therefore $fm_{1}=fm_{3}$. 4. (iv)

We assert that $m_{1}a=m_{3}a$. We have $m_{1}a_{1}=m_{1}$ due to $m_{1}(1_{R}-a_{1})=0$. Similarly, $m_{2}a_{2}=m_{2}$. From $m_{3}a=m_{3}a_{1}+m_{3}a_{2}a_{1}-m_{3}a_{1}a_{2}a_{1}$ and since $m_{3}a_{2}a_{1}=m_{2}a_{2}a_{1}=m_{2}a_{1}=m_{1}a_{1}=m_{1}$, we obtain

[TABLE]

We now claim that $a_{1}(1_{R}-a_{2})a_{1}=0$, i.e. $(1_{R}-a_{2})a_{1}\in r_{R}(a_{1})=r_{R}(m_{1})$. By $m_{1}a_{2}a_{1}=f_{1}m_{2}a_{2}a_{1}=f_{1}m_{2}a_{1}=m_{1}a_{1}$ it follows that $m_{1}(1_{R}-a_{2})a_{1}=0$. Then $a_{1}(1_{R}-a_{2})a_{1}=0$, as required. Hence $m_{3}a=m_{1}$ and thus $m_{1}a=m_{3}a$.

By Theorem 2.5, $m_{1}\leq^{-}m_{3}$ and so the relation $\leq^{-}$ is a partial order on $M$. ∎

We close this section by giving the star analogues of the minus order on modules. A ring equipped with an involution $\ast$ will be called a $\ast$-ring. A $\ast$-ring $R$ is called a Rickart $\ast$-ring if the left annihilator $l_{R}(a)$ of any element $a\in R$ is generated by a projection (i.e. a self-adjoint idempotent), equivalently, the right annihilator $r_{R}(a)$ of any element $a\in R$ is generated by a projection. Recall that Drazin [6] defined the star partial order in a general setting of proper $\ast$-semigroups and note that natural special cases of proper $\ast$-semigroups are all proper $\ast$-rings, with “properness” defined via $aa^{\ast}=0$ implying $a=0$. Recall also (see e.g. [2]) that any Rickart $\ast$-ring has the identity and is a proper $\ast$-ring.

For Rickart $\ast$-rings, the following characterization of the star partial order $\underset{\ast}{\leq}$ was given in [10]. Let $R$ be a Rickart $\ast$-ring. For $a,b\in R$ we have $a\underset{\ast}{\leq}b$ if and only if there exist projections $p,q\in R$ such that

[TABLE]

Observe here (see e.g. [4, Lemma 2.1]) that for any idempotent $p$ in a ring $R$ with identity,

[TABLE]

Two orders that are closely related to the star and the minus partial orders are the left-star and the right-star partial orders. These orders were introduced by Baksalary and Mitra in [1] on the set of $m\times n$ complex matrices $M_{m,n}(\mathbb{C})$. The left-star partial order is defined as follows. For $A,B\in M_{m,n}(\mathbb{C})$,

[TABLE]

The right star partial order ${\leq\ast}$ is defined symmetrically: $A{\leq\ast}\,B$ if and only if $AA^{\ast}=BA^{\ast}$ and $\mathrm{Im}\,A^{\ast}\subseteq\mathrm{Im}\,B^{\ast}$. These two orders were generalized in [5] from the set of all $n\times n$ complex matrices to $B(\mathcal{H})$ where $\mathcal{H}$ is an arbitrary complex Hilbert space. Using annihilators, authors further generalized in [10] the left and the right-star partial orders to Rickart $\ast$-rings.

Motivated (4) and (5), and by generalizations of the left-star and the right-star orders to Rickart $\ast$-rings in [10], we will now extend the notion of these orders to the module theoretic setting.

Definition 2.7**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. We write

• (i)

$m_{1}\leq\ast~{}m_{2}$ if there exist $f^{2}=f\in S$, $a^{2}=a=a^{*}\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$, where $R$ is a $\ast$-ring. We call the relation $\leq\ast$ the right-star order on $M$.

• (ii)

$m_{1}~{}\ast\leq~{}m_{2}$ if there exist $f^{2}=f=f^{*}\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$, where $S$ is a $\ast$-ring. We call the relation $\ast\leq$ the left-star order on $M$.

• (iii)

$m_{1}\underset{\ast}{\leq}m_{2}$ if there exist $f^{2}=f=f^{*}\in S$, $a^{2}=a=a^{*}\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$ and $m_{1}a=m_{2}a$, where $R$, and $S$ are $\ast$-rings. We call the relation $\underset{\ast}{\leq}$ the star order on $M$.

We will prove that the relations introduced with Definition 2.7 are partial orders when the module $M$ over a Rickart $*$-ring is regular.

Theorem 2.8**.**

Let $M$ be a regular module. Then the following hold.

1. (1)

If $R$ is a Rickart $\ast$-ring, then the relation $\leq\ast$, introduced with Definition 2.7, is a partial order on $M$. 2. (2)

If $S$ is a Rickart $\ast$-ring, then the relation $\ast\leq$, introduced with Definition 2.7, is a partial order on $M$. 3. (3)

If $R$ and $S$ are Rickart $\ast$-rings, then the relation $\underset{\ast}{\leq}$, introduced with Definition 2.7, is a partial order on $M$.

Proof.

We will only prove (1). The proofs of (2) and (3) are similar.

Reflexivity: Let $m\in M$. Since $M$ is regular, there exists $\varphi\in M^{*}$ such that $m=m\varphi m$. Let $f=m\varphi$ and $b=\varphi m$. Then $f^{2}=f\in S$ and $b^{2}=b\in R$. Clearly, $l_{S}(m)\subseteq l_{S}(f)$. From $m=m\varphi m$ we have $m=fm$. Thus $l_{S}(f)\subseteq l_{S}(m)$ and hence $l_{S}(m)=l_{S}(f)$. Similarly, $r_{R}(m)=r_{R}(b)$. By assumption $R$ is a Rickart $\ast$-ring so since $b\in R$, there exists (the unique) $a\in R$ with $a=a^{2}=a^{\ast}$ and $r_{R}(a)=r_{R}(b)$. To sum up, there exists $f^{2}=f\in S$, $a^{2}=a=a^{*}\in R$ such that $l_{S}(m)=l_{S}(f)$, $r_{R}(m)=r_{R}(a)$. We may conclude that $m\leq\ast~{}m$ for every $m\in M$.

Antisymmetry: Let $m_{1},m_{2}\in M$ with $m_{1}\leq\ast~{}m_{2}$ and $m_{2}\leq\ast~{}m_{1}$. Then there exist $f_{1}^{2}=f_{1}\in S$, $a_{1}^{2}=a_{1}=a_{1}^{*}\in R$ such that $l_{S}(m_{1})=l_{S}(f_{1})$, $r_{R}(m_{1})=r_{R}(a_{1})$, $f_{1}m_{1}=f_{1}m_{2}$, $m_{1}a_{1}=m_{2}a_{1}$; and $f_{2}^{2}=f_{2}\in S$, $a_{2}^{2}=a_{2}=a_{2}^{*}\in R$ such that $l_{S}(m_{2})=l_{S}(f_{2})$, $r_{R}(m_{2})=r_{R}(a_{2})$, $f_{2}m_{2}=f_{2}m_{1}$, $m_{2}a_{2}=m_{1}a_{2}$. By (5), $l_{S}(f_{1})=S(1_{S}-f_{1})$ and $r_{R}(a_{1})=(1_{R}-a_{1})R$ which imply by assumption that $(1_{S}-f_{1})m_{1}=0$ and $m_{1}(1_{R}-a_{1})=0$. So, $m_{1}=f_{1}m_{1}=m_{1}a_{1}=m_{2}a_{1}$. Similarly, $m_{2}=f_{2}m_{2}=f_{2}m_{1}$. Hence

[TABLE]

Transitivity: Let $m_{1},m_{2},m_{3}\in M$ with $m_{1}\leq\ast~{}m_{2}$ and $m_{2}\leq\ast~{}m_{3}$. Then there exist $f_{1}^{2}=f_{1}\in S$, $a_{1}^{2}=a_{1}=a_{1}^{*}\in R$ such that $l_{S}(m_{1})=l_{S}(f_{1})$, $r_{R}(m_{1})=r_{R}(a_{1})$, $f_{1}m_{1}=f_{1}m_{2}$, $m_{1}a_{1}=m_{2}a_{1}$; and $f_{2}^{2}=f_{2}\in S$, $a_{2}^{2}=a_{2}=a_{2}^{*}\in R$ such that $l_{S}(m_{2})=l_{S}(f_{2})$, $r_{R}(m_{2})=r_{R}(a_{2})$, $f_{2}m_{2}=f_{2}m_{3}$, $m_{2}a_{2}=m_{3}a_{2}$. From $(1_{S}-f_{1})m_{1}=0$ and $m_{1}(1_{R}-a_{1})=0$ we obtain

[TABLE]

Let $f=f_{1}f_{2}$ and $a=a_{1}$. To conclude the proof we will show that $f^{2}=f\in S$ and

$l_{S}(m_{1})=l_{S}(f)$, $fm_{1}=fm_{3}$, and $m_{1}a=m_{3}a$.

From $r_{R}(m_{2})=r_{R}(a_{2})$ we have $m_{2}=m_{2}a_{2}$ and thus $m_{1}a_{1}a_{2}=f_{1}m_{2}a_{2}=f_{1}m_{2}=m_{1}$. So $m_{1}(a_{1}a_{2}-1_{R})=0$. Since $r_{R}(m_{1})=r_{R}(a_{1})$, it follows that $a_{1}(a_{1}a_{2}-1_{R})=0$ and hence $a_{1}a_{2}=a_{1}$. Recall that $a_{1}$ and $a_{2}$ are self-adjoined. So, $a_{1}=a_{2}a_{1}$ which yields

[TABLE]

Since $l_{S}(m_{2})=l_{S}(f_{2})$, we have $m_{2}=f_{2}m_{2}$ which implies $f_{2}m_{1}=f_{2}m_{2}a_{1}=m_{2}a_{1}=m_{1}$. So, $(1_{S}-f_{1}f_{2})m_{1}=m_{1}-f_{1}f_{2}m_{1}=m_{1}-f_{1}m_{1}=m_{1}-m_{1}=0$. Since $l_{S}(m_{1})=l_{S}(f_{1})$, $(1-f_{1}f_{2})f_{1}=0$, so $f_{1}=f_{1}f_{2}f_{1}$. Now $fm_{1}=f_{1}f_{2}m_{1}=m_{1}$ and $fm_{3}=f_{1}f_{2}m_{3}=f_{1}f_{2}m_{2}=f_{1}m_{2}=m_{1}$. So

[TABLE]

Also $f^{2}=(f_{1}f_{2})^{2}=f_{1}f_{2}f_{1}f_{2}=f_{1}f_{2}$. Finally, since $f=f_{1}f_{2}$ and $f_{1}=f_{1}f_{2}f_{1}=ff_{1}$ we get, $l_{S}(m_{1})=l_{S}(f_{1})=l_{S}(f)$. We may conclude that $m_{1}\leq\ast~{}m_{3}$. ∎

3. Some Characterizations of the Minus order on Modules

In this section we investigate some properties and characterizations of the minus partial order on modules.

Proposition 3.1**.**

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

1. (1)

$m_{1}\leq^{-}m_{2}$ if and only if $gm_{1}\leq^{-}gm_{2}$ for every invertible element $g\in S$. 2. (2)

$m_{1}\leq^{-}m_{2}$ if and only if $m_{1}b\leq^{-}m_{2}b$ for every invertible element $b\in R$.

Proof.

(1) Assume first that $m_{1}\leq^{-}m_{2}$. By Theorem 2.5, there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$. Let $g\in S$ be invertible. Note first that $gm_{1}$ is a regular element in $M$ since $M$ is a regular module. Denote $p=gfg^{-1}$. Then $p\in S$ is an idempotent. We will show that $l_{S}(gm_{1})=l_{S}(p)$, $r_{R}(gm_{1})=r_{R}(a)$, $pgm_{1}=pgm_{2}$, and $gm_{1}a=gm_{2}a$.

1. (i)

If $h\in l_{S}(gm_{1})$, then $hgm_{1}=0$ and so $hgfm_{1}=0$. Hence $hgf\in l_{S}(m_{1})=l_{S}(f)$. Thus $hgf=0$ and so $hgfg^{-1}=0$. Therefore $l_{S}(gm_{1})\subseteq l_{S}(p)$. Conversely, if $x\in l_{S}(p)$, then $xp=xgfg^{-1}=0$. Hence $xgm_{1}=xgfm_{1}=xgfg^{-1}gm_{1}=0$ which yields $l_{S}(p)\subseteq l_{S}(gm_{1})$ and therefore $l_{S}(gm_{1})=l_{S}(p)$. 2. (ii)

If $x\in r_{R}(gm_{1})$, then $gm_{1}x=0$ and so $m_{1}x=0$. Hence $x\in r_{R}(m_{1})=r_{R}(a)$. Thus $r_{R}(gm_{1})\subseteq r_{R}(a)$. Conversely, if $x\in r_{R}(a)$, then $m_{1}x=0$ and so $gm_{1}x=0$. Hence $x\in r_{R}(gm_{1})$ and thus $r_{R}(a)\subseteq r_{R}(gm_{1})$. So, $r_{R}(gm_{1})=r_{R}(a)$. 3. (iii)

$pgm_{1}=gfg^{-1}gm_{1}=gfm_{1}=gfm_{2}=gfg^{-1}gm_{2}=pgm_{2}$. 4. (iv)

Since $m_{1}a=m_{2}a$, $gm_{1}a=gm_{2}a$.

Therefore $gm_{1}\leq^{-}gm_{2}$.

If $gm_{1}\leq^{-}gm_{2}$ for an invertible $g\in S$, then $g^{-1}gm_{1}\leq^{-}g^{-1}gm_{2}$, i.e. $m_{1}\leq^{-}m_{2}$.

(2) Assume first, that $m_{1}\leq^{-}m_{2}$. By Theorem 2.5, there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$. For an invertible element $b\in R$, let $q=b^{-1}ab$. Note again that $m_{1}b$ is a regular element in $M$ since $M$ is a regular module. Then $q\in R$ is an idempotent. We will show that $l_{S}(m_{1}b)=l_{S}(f)$, $r_{R}(m_{1}b)=r_{R}(q)$, $m_{1}bq=m_{2}bq$, and $fm_{1}b=fm_{2}b$.

1. (i)

If $h\in l_{S}(m_{1}b)$, then $hm_{1}b=0$ and so $hm_{1}bb^{-1}=0$. Hence $hm_{1}=0$ and so $h\in l_{S}(m_{1})=l_{S}(f)$. Thus $l_{S}(m_{1}b)\subseteq l_{S}(f)$. Conversely, if $h\in l_{S}(f)=l_{S}(m_{1})$, then $hm_{1}=0$ and so $hm_{1}b=0$. Hence $h\in l_{S}(m_{1}b)$, i.e. $l_{S}(f)\subseteq l_{S}(m_{1}b)$, and therefore $l_{S}(m_{1}b)=l_{S}(f)$. 2. (ii)

If $x\in r_{R}(q)$, then $qx=0$ and thus $b^{-1}abx=0$. Hence $bb^{-1}abx=0$ and so $abx=0$. From $bx\in r_{R}(a)=r_{R}(m_{1})$ we obtain $m_{1}bx=0$ and therefore $r_{R}(q)\subseteq r_{R}(m_{1}b)$. Conversely, if $x\in r_{R}(m_{1}b)$, then $m_{1}bx=0$ which yields $bx\in r_{R}(m_{1})=r_{R}(a)$. Hence $abx=0$ and so $b^{-1}abx=0$. Thus $qx=0$, i.e. $x\in r_{R}(q)$. It follows that $r_{R}(m_{1}b)\subseteq r_{R}(q)$ and therefore $r_{R}(m_{1}b)=r_{R}(q)$. 3. (iii)

$m_{1}bq=m_{1}bb^{-1}ab=m_{1}ab=m_{2}ab=m_{2}bb^{-1}ab=m_{2}bq$. 4. (iv)

Since $fm_{1}=fm_{2}$, $fm_{1}b=fm_{2}b$.

We conclude that $m_{1}b\leq^{-}m_{2}b$.

If $m_{1}b\leq^{-}m_{2}b$ for an invertible $b\in R$, then $m_{1}bb^{-1}\leq^{-}m_{2}bb^{-1}$, i.e. $m_{1}\leq^{-}m_{2}$. ∎

We will next prove that if we somewhat relax conditions on left and right annihilators in Theorem 2.5, we still obtain the minus partial order on regular modules.

Theorem 3.2**.**

Let $M$ be a regular module and $m_{1},m_{2}\in M$. Then $m_{1}\leq^{-}m_{2}$ if and only if there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(f)\subseteq l_{S}(m_{1})$, $r_{R}(a)\subseteq r_{R}(m_{1})$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$.

Proof.

Suppose first that there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(f)\subseteq l_{S}(m_{1})$, $r_{R}(a)\subseteq r_{R}(m_{1})$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$. Since $M$ is a regular module, for $m_{1}\in M$, there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$. Let

[TABLE]

Clearly, $f_{1}^{2}=f_{1}$ and $a_{1}^{2}=a_{1}$. We will first show that $l_{S}(f_{1})=l_{S}(m_{1})$ and $r_{R}(a_{1})=r_{R}(m_{1})$. Let $g\in l_{S}(m_{1})$. Then $gm_{1}=0$ and so $gf_{1}=gm_{1}\varphi=0$. Hence $g\in l_{S}(f_{1})$ and thus $l_{S}(m_{1})\subseteq l_{S}(f_{1})$. Conversely, if $g\in l_{S}(f_{1})$, then $gf_{1}=0$ and so $gm_{1}=gm_{1}\varphi m_{1}=gf_{1}m_{1}=0$. It follows that $g\in l_{S}(m_{1})$, i.e. $l_{S}(f_{1})\subseteq l_{S}(m_{1})$, and thus $l_{S}(f_{1})=l_{S}(m_{1})$. Now let $b\in r_{R}(m_{1})$. Then $m_{1}b=0$ and so $a_{1}b=\varphi m_{1}b=0$. Hence $b\in r_{R}(a_{1})$ and so $r_{R}(m_{1})\subseteq r_{R}(a_{1})$. For the reverse inclusion, let $b\in r_{R}(a_{1})$. Then $m_{1}b=m_{1}\varphi m_{1}b=m_{1}a_{1}b=0$ and hence $b\in r_{R}(m_{1})$. So, $r_{R}(m_{1})=r_{R}(a_{1})$.

Next we will prove that $l_{S}(f)\subseteq l_{S}(f_{1})$ and $r_{R}(a)\subseteq r_{R}(a_{1})$. Let $g\in l_{S}(f)$. Then $gf=0$ and so $gf_{1}=gm_{1}\varphi=gfm_{1}\varphi=0$. Hence $g\in l_{S}(f_{1})$, i.e. $l_{S}(f)\subseteq l_{S}(f_{1})$. For the other inclusion, let $b\in r_{R}(a)$. Then $ab=0$ and so $a_{1}b=\varphi m_{1}b=\varphi m_{1}ab=0$. It follows that $b\in r_{R}(a_{1})$, i.e. $r_{R}(a)\subseteq r_{R}(a_{1})$. Since then $(1_{S}-f)f_{1}=0$ and $a_{1}(1_{R}-a)=0$ and since $f_{1}$ and $a_{1}$ are idempotents, we have $f_{1}=ff_{1}=f_{1}ff_{1}$ and $a_{1}=a_{1}a=a_{1}aa_{1}$. Note that $f_{1}f\in S$ and $aa_{1}\in R$ are idempotents.

We will now prove that $l_{S}(m_{1})=l_{S}(f_{1}f)$, $r_{R}(m_{1})=r_{R}(aa_{1})$. Clearly, $l_{S}(m_{1})=l_{S}(f_{1})\subseteq l_{S}(f_{1}f)$ and $r_{R}(m_{1})=r_{R}(a_{1})\subseteq r_{R}(aa_{1})$. For any $x\in l_{S}(f_{1}f)$ and any $y\in r_{R}(aa_{1})$, we have $xf_{1}=xf_{1}ff_{1}=0$ and $a_{1}y=a_{1}aa_{1}y=0$. This yields $x\in l_{S}(f_{1})=l_{S}(m_{1})$ and $y\in r_{R}(a_{1})=r_{R}(m_{1})$. It follows that $l_{S}(m_{1})=l_{S}(f_{1}f)$ and $r_{R}(m_{1})=r_{R}(aa_{1})$, as desired.

Finally, from $fm_{1}=fm_{2}$ and $m_{1}a=m_{2}a$, we establish that $f_{1}fm_{1}=f_{1}fm_{2}$ and $m_{1}aa_{1}=m_{2}aa_{1}$. Therefore $m_{1}\leq^{-}m_{2}$.

The converse implication follows directly by Theorem 2.5. ∎

Corollary 3.3**.**

Let $M$ be a regular module and $m_{1},m_{2}\in M$. Then $m_{1}\leq^{-}m_{2}$ if and only if there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $m_{1}R\subseteq fM$, $Sm_{1}\subseteq Ma$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$.

Proof.

Note that for any $m\in M$, $f^{2}=f\in S$ and $a^{2}=a\in R$,

1. (1)

$l_{S}(f)\subseteq l_{S}(m)$ if and only if $mR\subseteq fM$, 2. (2)

$r_{R}(a)\subseteq r_{R}(m)$ if and only if $Sm\subseteq Ma$.

The result is thus an immediate consequence of Theorem 3.2. ∎

There are many definitions that are equivalent to Hartwig’s definition of the minus partial order. One of the most commonly used is the following which is due to Jones (see for example [8]). Let $\mathcal{S}$ be a semigroup. Then

[TABLE]

where $p,q$ are idempotent elements in $\mathcal{S}^{1}$ and $\mathcal{S}^{1}$ denotes $\mathcal{S}$ if $\mathcal{S}$ has the identity, and $\mathcal{S}$ with the identity adjoined in the other case. Jones introduced this relation for the setting of an arbitrary semigroup. If $\mathcal{S}$ is a regular semigroup, then it is known (see [11]) that $\leq_{J}$ is a partial order on $\mathcal{S}$. We will now introduce a new relation on a module $M$ that is analogous to (6) and then show that when $M$ is a regular module this new relation is the minus partial order on $M$.

Definition 3.4**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. We write $m_{1}\leq_{J}m_{2}$ if there exist idempotent elements $f\in S$ and $a\in R$ such that $m_{1}=fm_{2}=m_{2}a$.

Lemma 3.5**.**

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 $m_{1}=fm_{1}=fm_{2}=m_{1}a=m_{2}a$.

Proof.

If $m_{1}\leq^{-}m_{2}$, then there exist $f^{2}=f\in S$, $a^{2}=a\in R$ such that $l_{S}(m_{1})=l_{S}(f)$, $r_{R}(m_{1})=r_{R}(a)$, $fm_{1}=fm_{2}$, and $m_{1}a=m_{2}a$. Hence $(1_{S}-f)m_{1}=0$ and $m_{1}(1_{R}-a)=0$, and therefore $m_{1}=fm_{1}=fm_{2}=m_{1}a=m_{2}a$. ∎

Theorem 3.6**.**

Let $M$ be a regular module and $m_{1},m_{2}\in M$. Then $m_{1}\leq_{J}m_{2}$ if and only if $m_{1}\leq^{-}m_{2}$.

Proof.

If $m_{1}\leq^{-}m_{2}$, then $m_{1}\leq_{J}m_{2}$ by Lemma 3.5. Conversely, let $m_{1}\leq_{J}m_{2}$. Then $m_{1}=fm_{2}=m_{2}a$ for some $f^{2}=f\in S$, $a^{2}=a\in R$. Clearly, $fm_{1}=fm_{2}$ and $m_{1}a=m_{2}a$. From $m_{1}=fm_{2}$ we get $l_{S}(f)\subseteq l_{S}(m_{1})$ and by $m_{1}=m_{2}a$ we obtain $r_{R}(a)\subseteq r_{R}(m_{1})$. Therefore an application of Theorem 3.2 completes the proof. ∎

In [11], Mitsch generalized Hartwig’s definition of the minus partial order to arbitrary semigroups. The definition follows: Suppose $a,b$ are two elements of an arbitrary semigroup $\mathcal{S}$. Then we write

[TABLE]

for some elements $x,y\in\mathcal{S}^{1}$. Mitsch proved that $\leq_{M}$ is indeed a partial order for any semigroup $\mathcal{S}$ and that for $a,b\in\mathcal{S}$,

[TABLE]

for some elements $x,y\in\mathcal{S}^{1}$. Following the above ideas, we now introduce a new relation $\leq_{M}$ on a module $M$.

Definition 3.7**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. We write $m_{1}\leq_{M}m_{2}$ if there exist $f\in S$ and $a\in R$ such that $m_{1}=m_{2}a=fm_{2}$ and $m_{1}=fm_{1}$.

We will show that when $M$ is a regular module, $\leq_{M}$ is again the minus partial order on $M$. First, we present an auxiliary result which was inspired by (8).

Lemma 3.8**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. Then $m_{1}\leq_{M}m_{2}$ if and only if there exist $f\in S$ and $a\in R$ such that

[TABLE]

Proof.

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 $m_{1}=fm_{1}$. Hence $m_{1}a=(fm_{2})a=f(m_{2}a)=fm_{1}=m_{1}$. The converse is obvious. ∎

Theorem 3.9**.**

Let $M$ be a regular module and $m_{1},m_{2}\in M$. Then $m_{1}\leq^{-}m_{2}$ if and only if $m_{1}\leq_{M}m_{2}$.

Proof.

Let $m_{1}\leq^{-}m_{2}$. Then by Theorem 3.6 there exist $f^{2}=f\in S$ and $a^{2}=a\in R$ with $m_{1}=fm_{2}=m_{2}a$. It follows, $(1_{S}-f)m_{1}=0$ and so $m_{1}=fm_{1}$. Hence $m_{1}\leq_{M}m_{2}$.

Conversely, let $m_{1}\leq_{M}m_{2}$. Then there exist $f\in S$ and $a\in R$ with $m_{1}=fm_{2}=m_{2}a$ and $m_{1}=fm_{1}$. Since $M$ is a regular module, there exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$. Let $g=m_{1}\varphi\in S$ and $b=\varphi m_{1}\in R$. Obviously, $g^{2}=g$ and $b^{2}=b$. Also, clearly, $l_{S}(m_{1})=l_{S}(g)$ and $r_{R}(m_{1})=r_{R}(b)$. Equation $m_{1}=fm_{1}$ implies $(1_{S}-f)m_{1}=0$. Therefore $(1_{S}-f)g=0$ and hence $g=fg$. It follows $(gf)^{2}=gf$. Also, $m_{1}a=m_{1}$ by Lemma 3.8. Then $m_{1}(1_{R}-a)=0$ and so $b(1_{R}-a)=0$ which implies $b=ba$. Hence $(ab)^{2}=ab$. We now take into account the following items:

1. (i)

Let $x\in l_{S}(gf)$. Then $0=xgf=xm_{1}\varphi f$ and so

[TABLE]

Hence $x\in l_{S}(m_{1})$ and thus $l_{S}(gf)\subseteq l_{S}(m_{1})$. 2. (ii)

Let $x\in r_{R}(ab)$. Then $0=abx=a\phi m_{1}x$ and so

[TABLE]

Hence $x\in r_{R}(m_{1})$ and thus $r_{R}(ab)\subseteq r_{R}(m_{1})$. 3. (iii)

$gfm_{1}=gm_{1}=gfm_{2}$. 4. (iv)

$m_{1}ab=m_{1}b=m_{2}ab$.

By Theorem 3.2, we may conclude that $m_{1}\leq^{-}m_{2}$. ∎

Corollary 3.10**.**

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

1. (i)

$m_{1}\leq^{-}m_{2}$; 2. (ii)

There exist $g\in S$ and $b\in R$ such that $m_{1}=m_{2}b=gm_{2}$ and $m_{1}=m_{1}b$.

Proof.

(i) $\Rightarrow$ (ii) Clear by Theorem 3.9 and Lemma 3.8.

(ii) $\Rightarrow$ (i) There exist $g\in S$ and $b\in R$ such that $m_{1}=m_{2}b=gm_{2}$ and $m_{1}=m_{1}b$. We have

[TABLE]

and thus (i) holds. ∎

With the next result, which follows directly by Lemma 3.5, we will present a relation between annihilators of elements $m_{1}$ and $m_{2}$ of a module $M$ where $m_{1}$ and $m_{2}$ are connected by the minus partial order.

Proposition 3.11**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. If $m_{1}\leq^{-}m_{2}$, then $l_{S}(m_{2})\subseteq l_{S}(m_{1})$ and $r_{R}(m_{2})\subseteq r_{R}(m_{1})$.

The following example shows that the converse statement of Proposition 3.11 need not hold in general.

Example 3.12**.**

Let $\mathbb{Z}_{10}$ denote the ring of integers modulo $10$. It is known that End${}_{\mathbb{Z}}(\mathbb{Z}_{10})\cong\mathbb{Z}_{10}$. Since $\mathbb{Z}_{10}$ is a semisimple $\mathbb{Z}$-module, it is a regular module. Note that the nontrivial idempotents of $\mathbb{Z}_{10}$ are $\overline{5}$ and $\overline{6}$. Consider the elements $m_{1}=\overline{2}$ and $m_{2}=\overline{6}$ of $\mathbb{Z}_{10}$. Then, on the one hand, $l_{{}_{\mathbb{Z}_{10}}}(m_{1})=l_{{}_{\mathbb{Z}_{10}}}(m_{2})=\{\overline{0},\overline{5}\}$ and $r_{{}_{\mathbb{Z}}}(m_{1})=r_{{}_{\mathbb{Z}}}(m_{2})=5\mathbb{Z}$. On the other hand, there are no idempotents $f\in$ End${}_{\mathbb{Z}}(\mathbb{Z}_{10})$ and $a\in\mathbb{Z}$ such that $m_{1}=fm_{2}=m_{2}a$. Thus, $m_{1}\not\leq^{-}m_{2}$. **

In [3], Blackwood at al. defined a partial order $\leq^{\oplus}$ in the following way:

[TABLE]

and called it the direct sum partial order. The direct sum partial order is known to be equivalent to the minus partial order on von Neumann regular rings (see [3, Lemma 3]). Motivated by the notion of the direct sum partial order on (regular) rings, we will conclude the paper by introducing a new relation on a module $M$ and then show that this relation is the minus partial order when the module $M$ is regular.

Definition 3.13**.**

Let $M$ be a module and $m_{1},m_{2}\in M$. We write

[TABLE]

Theorem 3.14**.**

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

1. (i)

$m_{1}\leq^{-}m_{2}$; 2. (ii)

$m_{1}\leq^{\oplus}m_{2}$.

Proof.

$(1)\Rightarrow(2)$: Let $m_{1}\leq^{-}m_{2}$. There exists $\varphi\in M^{*}$ such that $m_{1}=m_{1}\varphi m_{1}$. By Remark 2.2, we have $M=m_{1}R\oplus T$ where $T=\{x\in M~{}|~{}m_{1}\varphi x=0\}$. To prove $m_{1}\leq^{\oplus}m_{2}$, we will show that $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$. Let first $x\in m_{2}R$. Then $x=m_{2}r$ for some $r\in R$ and therefore

[TABLE]

By Proposition 2.4, $m_{1}R\subseteq m_{2}R$ which yields $(m_{2}-m_{1})R\subseteq m_{2}R$. Thus,

[TABLE]

Let $a\in R.$ Observe that

[TABLE]

and therefore $(m_{2}-m_{1})R\subseteq T$. Since $M=m_{1}R\oplus T$, it follows that $m_{1}R\cap(m_{2}-m_{1})R=\{0\}.$ Thus, $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$.

(2) $\Rightarrow$ (1): Let $m_{1}\leq^{\oplus}m_{2}$, i.e. $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$. There exist $\varphi_{2}\in M^{*}$ and $r\in R$ with $m_{2}=m_{2}\varphi_{2}m_{2}$ and $m_{1}=m_{2}r$. Then $m_{2}\varphi_{2}m_{1}=m_{2}\varphi_{2}m_{2}r=m_{2}r=m_{1}$ and hence

[TABLE]

Thus $m_{1}\varphi_{2}m_{1}-m_{1}=-(m_{2}-m_{1})\varphi_{2}m_{1}\in m_{1}R\cap(m_{2}-m_{1})R$. Since $m_{1}R\cap(m_{2}-m_{1})R=\{0\}$, it follows $m_{1}\varphi_{2}m_{1}=m_{1}$. Therefore $m_{2}\varphi_{2}m_{1}=m_{1}=m_{1}\varphi_{2}m_{1}$. Let $x=\varphi_{2}m_{1}\varphi_{2}\in M^{*}$. Then $m_{1}xm_{1}=m_{1}$. Note that $m_{2}x=m_{2}\varphi_{2}m_{1}\varphi_{2}=m_{1}\varphi_{2}$ and $m_{1}x=m_{1}\varphi_{2}m_{1}\varphi_{2}=m_{1}\varphi_{2}$ and hence $m_{1}x=m_{2}x$. Also, $m_{2}\varphi_{2}(m_{2}-m_{1})=m_{2}\varphi_{2}m_{2}-m_{2}\varphi_{2}m_{1}=m_{2}-m_{1}$, so

[TABLE]

and this implies that $m_{1}\varphi_{2}(m_{2}-m_{1})=(m_{2}-m_{1})-(m_{2}-m_{1})\varphi_{2}(m_{2}-m_{1})\in m_{1}R\cap(m_{2}-m_{1})R$. Since $m_{1}R\cap(m_{2}-m_{1})R=0$, we obtain $m_{1}\varphi_{2}m_{2}=m_{1}\varphi_{2}m_{1}$ and $m_{2}-m_{1}=(m_{2}-m_{1})\varphi_{2}(m_{2}-m_{1})$. From $m_{1}\varphi_{2}m_{1}=m_{1}$, it follows $m_{1}\varphi_{2}m_{2}=m_{1}$. Now, $xm_{2}=\varphi_{2}m_{1}\varphi_{2}m_{2}=\varphi_{2}m_{1}$ and $xm_{1}=\varphi_{2}m_{1}\varphi_{2}m_{1}=\varphi_{2}m_{1}$. Thus $xm_{1}=xm_{2}$ and therefore $m_{1}\leq^{-}m_{2}$. ∎

The following result follows directly from Theorem 3.14.

Corollary 3.15**.**

If $M$ is a regular module, then Definitions 2.3 and 3.13 are equivalent, i.e. the relation $\leq^{\oplus}$ is the minus partial order on $M$.

It can be seen that for nonzero elements, the minus order is the identity relation on vector spaces. But this is not the case if the module is regular as the following example shows.

Example 3.16**.**

Consider $M=\mathbb{Z}_{6}$ as an $R=\mathbb{Z}_{30}$-module. On the one hand, since $R\cong M\oplus\mathbb{Z}_{5}$, $M$ is a projective $R$-module. On the other hand, $M$ is semisimple as an $R$-module. Hence $M$ is a regular $R$-module. Let $m_{1}=\overline{2}$, $m_{2}=\overline{5}\in M$. Then $m_{2}R=m_{1}R\oplus(m_{2}-m_{1})R$, i.e., $\overline{5}\mathbb{Z}_{30}=M=\overline{2}\mathbb{Z}_{30}\oplus\overline{3}\mathbb{Z}_{30}$. This implies $m_{1}\leq^{\oplus}m_{2}$. Therefore $m_{1}\leq^{-}m_{2}$ by Theorem 3.14.**

Acknowledgment. The work is partially supported by Slovene Research Agency, ARRS (Slovene-Turkish Grant BI-TR/17-19-004) and by The Scientific and Technological Research Council of Turkey, TUBITAK (Grant TUBITAK-116F435). The authors wish to thank ARRS and TUBITAK for financial support.