Co-EP-Ness and EP-Ness Involving the Inverse along an Element

Abstract

Let $a,d$ be two elements in rings and $a^{\parallel d}$ be the inverse of a along d. When $a^{\parallel d}$ exists, we obtain several characterizations for the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$, which is related to the invertibility of elements expressed by certain functions of $a,d$ and suitable elements from the center of the ring. On the other hand, some equivalent conditions for the equality $a{a}^{\parallel d}=a^{\parallel d}a$, as the complement of the previous invertibility in some sense, are given by means of the group inverses and the ring units, respectively. Then, the results obtained are applied in a $\ast$-ring, namely, when $d=a^{\ast }$, the co-EP and EP properties are deduced correspondingly.

1. Introduction

Throughout this paper, R denotes an associative ring with unity 1, and $\mathbb{N}$ means the set of all positive integers. Denote by $C\left(R\right)$ the center of R, that is, the set of such elements that commute with all elements of R. The right and left annihilator of $a\in R$ are defined by $a^0=\{x\in R|ax=0\}$ and ${}^{0}a=\{x\in R|xa=0\}$, respectively. An involution ∗: R→R is an anti-isomorphism: ${\left(a^{\ast }\right)}^{\ast }=a$, ${(a+b)}^{\ast }=a^{\ast }+b^{\ast }$ and ${\left(ab\right)}^{\ast }=b^{\ast }{a}^{\ast }$ for all $a,b\in R$. We call R a ∗-ring if there exists an involution ∗ on R. In a ∗-ring R, an element $a\in R$ is left (resp. right) ∗-cancelable if $a^{\ast }ax=a^{\ast }ay$ (resp. $xa{a}^{\ast }=ya{a}^{\ast }$) implies $ax=ay$ (resp. $xa=ya$). Then, $a\in R$ is ∗-cancelable if it is both left and right ∗-cancelable.

For the readers’ convenience, we first recall the definitions of some types of generalized inverses. An element $a\in R$ is said to be Moore–Penrose invertible with respect to the involution ∗ [1] if the following equations

$$\left(1\right)axa=a,\left(2\right)xax=x,\left(3\right){\left(ax\right)}^{\ast }=ax,\left(4\right){\left(xa\right)}^{\ast }=xa$$

have a common solution. Such solution is unique if it exists, and is denoted by $a^{†}$. An element a is regular if there exists x satisfying the Equation (1), such particular solution x is denoted by $a^{-}$. If $x\in R$ satisfies both Equations (1) and (3), then x is called a $\{1,3\}$-inverse of a and denoted by $a^{(1,3)}$. The set of all $\{1,3\}$-invertible elements of R is denoted by $R^{\{1,3\}}$. Similarly, if $x\in R$ satisfies both Equations (1) and (4), then x is called a $\{1,4\}$-inverse of a and denoted by $a^{(1,4)}$. The set of all $\{1,4\}$-invertible elements of R is denoted by $R^{\{1,4\}}$. As is well known, if $a\in R$ is Moore–Penrose invertible, then a is ∗-cancelable [2].

The Drazin inverse [3] of $a\in R$ is the element $x\in R$ which satisfies

$$xax=x,ax=xa\mathrm{and}{a}^k=a^{k+1}x,\mathrm{for}\mathrm{some}k\in \mathbb{N}.$$

The element x is unique if it exists and is denoted by $a^D$. Particularly, if ${\mathit{aa}}^{D}a=a$, then the Drazin inverse $a^D$ is called the group inverse of a, and it is denoted by $a^{\#}$.

In 2011, Mary [4] defined a new generalized inverse, called the inverse along an element in a ring or semigroup, which covers the usual concept of unit and the above classic generalized inverses. The element $a\in R$ is said to be invertible along $d\in R$ [4] if there exists $b\in R$ such that

$$bad=d=dab, bR\subseteq dR \mathrm{and} Rb\subseteq Rd$$

i.e.,

$$bab=b, bR=dR \mathrm{and} Rb=Rd.$$

If such b exists, then it is unique and is said to be the inverse of a along d, which is denoted by $a^{\parallel d}$. In particular, $a^{\parallel 1}=a^{-1}$, $a^{\parallel a}=a^{\#}$ and $a^{\parallel a^{\ast }}=a^{†}$. Moreover, if $a{a}^{\parallel d}a=a$, then we say that $a^{\parallel d}$ is an inner inverse of a along d, and a is inner invertible along d.

The symbols $R^{\parallel d}$, $R^{\parallel •d}$, $R^{†}$, $R^{\#}$, $R^{-1}$ and $R^{-}$ mean the sets of all invertible along d, inner invertible along d, Moore–Penrose invertible, group invertible, invertible and regular elements of the ring R, respectively.

This whole story began with the notion of EP matrix $A\in {\mathbb{C}}^{n\times n}$ [5], that is $\mathcal{R}\left(A\right)=\mathcal{R}\left(A^{\ast }\right)$, i.e., $A{A}^{†}=A^{†}A$, where $\mathcal{R}\left(A\right)$ denotes the column space of A. This was followed by further research in different settings, such as complex matrices, operators on a Hilbert space, $C^{\ast }$-algebras and ∗-rings. In a ∗-ring R, naturally, an element $a\in R$ is said to be an EP element if $a\in R^{†}$ and $a{a}^{†}=a^{†}a$. We use $R^{\mathrm{EP}}$ to stand for the sets of all EP elements of R. Koliha and Patrício [2] proved that $a\in R^{\mathrm{EP}}$ if and only if $a\in R^{\#}$ and $a{a}^{\#}={\left(a{a}^{\#}\right)}^{\ast }$. Later, Patrício and Puystjens [6] deduced that $a\in R^{\mathrm{EP}}$ if and only if ($a\in R^{\#}$ and $aR=a^{\ast }R$) if and only if ($a\in R^{†}$ and $aR=a^{\ast }R$). In [7], Mosić and Djordjević characterized the EP element by $a\in R^{\#}\cap R^{†}$ and ${\left(a^{\#}\right)}^n={\left(a^{†}\right)}^n$, where $n\in \mathbb{N}$. More results about this topic can be found in [8,9].

Based on the EP matrix, another class of matrices called co-EP was introduced by Benítez and Rakočević [10]. A square matrix A is said to be co-EP if $A{A}^{†}-A^{†}A$ is invertible. In some sense, the class of co-EP matrices is the complement of the class of EP matrices. It is interesting that if A is a co-EP matrix, then

$$A-A^{\ast },A{A}^{\ast }+A^{\ast }A\mathrm{and}A{A}^{†}+A^{†}A,$$

are all nonsingular matrices [10]. In 2021, Zuo, Baksalary and Cvetković-Ilić [11] proved that for $a,b,c\in \mathbb{C}$ and $ab\ne 0$, the following matrices

$$aA+b{A}^{†}+cA{A}^{†},aA+b{A}^{\ast }+cA{A}^{\ast }\mathrm{and}aA{A}^{†}+b{A}^{†}A+cA{\left(A^{†}\right)}^{2}A,$$

are also nonsingular matrices, under the condition that A is a co-EP matrix. Recently, for $a,b,c,d,\lambda ,\mu ,\xi ,\eta \in \mathbb{C}$, $ab\ne cd$ and $\mu \xi \ne 0$, the invertibility of the following linear combinations

$$aA+b{A}^{†}+cA{A}^{†}+d{A}^{†}A\mathrm{and}\lambda A^{†}+\mu A{A}^{†}+\xi A^{†}A+\eta A{\left(A^{†}\right)}^{2}A,$$

was proved, when A is a co-EP matrix [12]. For the ∗-ring case, Benítez, Liu and Rakočević [13] studied some characterizations for the co-EP element, namely $a\in R^{†}$ and $a{a}^{†}-a^{†}a$ is invertible. In addition, they investigated more general cases, that is, the invertibility of $ab-ba$, where $aba=a$ and $bab=b$. This invertibility was also considered in [12]. More results about co-EP properties can be seen in [8,10,14,15,16,17].

As we all know that the Moore–Penrose inverse belongs to the inverse along an element, in this paper, we consider generalizations of the co-EP and EP properties by means of the inverse along an element in two directions. One is the characterizations for the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$, when $a\in R^{\parallel d}$. We show that this invertibility can imply the invertibility of ${\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}a{d}^{2}a+{\lambda }_{4}d{a}^{2}d$, where ${\lambda }_i\in C\left(R\right)$, $i=\overline{1,4}$. Furthermore, when $a\in R^{\parallel •d}$, the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$ is related to that of ${\delta }_{1}a+{\delta }_{2}d+{\delta }_{3}ad+{\delta }_{4}da$, where ${\delta }_i\in C\left(R\right)$, $i=\overline{1,4}$. In particular, when both a and d are idempotents, such invertibility is also investigated. The other is the characterizations for the equality $a{a}^{\parallel d}=a^{\parallel d}a$, when $d\in R^{\parallel a}$. Some equivalent conditions for $a{a}^{\parallel d}=a^{\parallel d}a$ are given by means of the group inverses and the ring units, respectively. As applications, some existence criteria of co-EP and EP elements are directly obtained.

2. Preliminaries

In this section, we start with some existence criteria for the inverse along an element.

Lemma 1

([18] Theorem 2.1). Let $a,d\in R$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$.

(ii)

$dR\subseteq daR$ and $da\in R^{\#}$.

(iii)

$Rd\subseteq Rad$ and $ad\in R^{\#}$.

In this case, $a^{\parallel d}=d{\left(ad\right)}^{\#}={\left(da\right)}^{\#}d$.

Lemma 2

([18] Theorem 2.2). Let $a,d\in R$. Then, $a\in R^{\parallel d}$ if and only if $Rd=Rdad$ and $dR=dadR$.

Applying Lemmas 1 and 2, we obtain

Lemma 3.

Let $a,d\in R$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel •d}$.

(ii)

$d\in R^{\parallel •a}$.

(iii)

$a\in R^{\parallel d}$ and $d\in R^{\parallel a}$.

In this case, $a{a}^{\parallel d}=d^{\parallel a}d$ and $a^{\parallel d}a=d{d}^{\parallel a}$.

Proof. 

(i) ⇔ (ii). Suppose that $a\in R^{\parallel •d}$. Applying Lemma 1, we obtain

$$a=a{a}^{\parallel d}a=a{\left(da\right)}^{\#}da=\left(ada\right){\left({\left(da\right)}^{\#}\right)}^{3}d\left(ada\right)\in adaR\cap Rada.$$

Hence, by Lemma 2, we obtain $d\in R^{\parallel a}$ and $d^{\parallel a}=a{\left(da\right)}^{\#}={\left(ad\right)}^{\#}a$. In addition, we conclude $a{a}^{\parallel d}=a{\left(da\right)}^{\#}d=d^{\parallel a}d$ and $a^{\parallel d}a=d{\left(ad\right)}^{\#}a=d{d}^{\parallel a}$. Then, $d{d}^{\parallel a}d=a^{\parallel d}ad=d$, so $d\in R^{\parallel •a}$.

(i) ⇒ (iii) is trivial by the implication (i) ⇒ (ii).

(iii) ⇒ (i). Using the hypotheses, we obtain $a{a}^{\parallel d}a=a{\left(da\right)}^{\#}da=d^{\parallel a}da=a$. Hence, $a\in R^{\parallel •d}$. □

Remark that $a\in R^{\parallel d}$ does not imply $d\in R^{\parallel a}$ in general. For example, let $R={\mathbb{C}}^{2\times 2}$, $d=0$ and $a=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, clearly $a\in R^{\parallel d}$, but $d\in R^{\parallel a}$. The next lemma gives some equivalent conditions for the invertibility of the difference of idempotents, which plays an important role in the sequel.

Lemma 4

([13] Theorem 1 and [19] Theorem 3.2). Let  $f,g\in R$ be idempotents. Then, the following statements are equivalent:

(i)

$f-g\in R^{-1}$.

(ii)

$fR\oplus gR=R$ and $Rf\oplus Rg=R$.

(iii)

There exist idempotents $h,k\in R$ such that $fh=h$, $hf=f$, $g(1-h)=1-h$, $(1-h)g=g$, $kf=k$, $fk=f$, $(1-k)g=1-k$ and $g(1-k)=g$.

(iv)

$1-fg\in R^{-1}$ and $f+g-fg\in R^{-1}$.

(v)

$f+g-fg\in R^{-1}$, $fR\cap gR=\left\{0\right\}$ and $f^0+g^0=R$.

As an application of Lemma 4, we obtain the following result.

Lemma 5.

Let $a,d\in R$ be such that $a\in R^{\parallel d}$. If $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$, then there exist idempotents $h,k\in R$ satisfying

$$\begin{array}{c}\hfill \begin{array}{cc}a{a}^{\parallel d}h=h, had=ad(i.e.,ha{a}^{\parallel d}=a{a}^{\parallel d}),a^{\parallel d}a(1-h)=1-h, hd=0(i.e.,h{a}^{\parallel d}a=0),& \hfill \\ and& (\ast )\hfill \\ ka{a}^{\parallel d}=k, dk=d(i.e.,a{a}^{\parallel d}k=a{a}^{\parallel d}), (1-k)a^{\parallel d}a=1-k, dak=0(i.e.,a^{\parallel d}ak=0).& \hfill \end{array}\end{array}$$

Proof. 

Since $a^{\parallel d}a{a}^{\parallel d}=a^{\parallel d}$, we have that both $a{a}^{\parallel d}$ and $a^{\parallel d}a$ are idempotents. By Lemma 4 (i) and (iii), there exist idempotents $h,k\in R$ such that

$$a{a}^{\parallel d}h=h, ha{a}^{\parallel d}=a{a}^{\parallel d}, a^{\parallel d}a(1-h)=1-h, h{a}^{\parallel d}a=0,$$

and

$$ka{a}^{\parallel d}=k, a{a}^{\parallel d}k=a{a}^{\parallel d}, (1-k)a^{\parallel d}a=1-k, a^{\parallel d}ak=0.$$

Note that $a^{\parallel d}ad=d=da{a}^{\parallel d}$. So, we get $had=\left(ha{a}^{\parallel d}\right)ad=a{a}^{\parallel d}ad=ad$ and $hd=\left(h{a}^{\parallel d}a\right)d=0$. Similarly, we have $dk=d$ and $dak=0$. □

Lemma 6.

(Jacobson’s lemma) Let $a,b\in R$. Then, $1-ab\in R^{-1}$ if and only if $1-ba\in R^{-1}$.

Lemma 7

([20] Theorem 1). Let $a\in R$. Then $a\in R^{\#}$ if and only if $a\in a^{2}R\cap R{a}^2$. In this case, if $a=a^{2}x=y{a}^2$, then $a^{\#}=a{x}^2=y^{2}a=yax$.

Lemma 8

([21] p. 201). Let R be a ∗-ring and $a\in R$. Then,

(i)

$a\in R^{\{1,3\}}$ if and only if $a=x^{\ast }{a}^{\ast }a$ for some $x\in R$. In this case, x is a $\{1,3\}$-inverse of a.

(ii)

$a\in R^{\{1,4\}}$ if and only if $a=a{a}^{\ast }{y}^{\ast }$ for some $x\in R$. In this case, y is a $\{1,4\}$-inverse of a.

Let $R_{\mathrm{co}}^{\mathrm{EP}}$ stand for the set of all co-EP elements of R. Recall that the Jacobson radical of R is defined by

$$J\left(R\right)=\{a\in R|1+ax\in R^{-1},foranyx\in R\}.$$

Note that $J\left(R\right)$ is a two-sided ideal and $\overline{R}:=R/J\left(R\right)$ is a quotient ring. For $a\in R$ and $A\subseteq R$, we use $\overline{a}$ and $\overline{A}$ to denote the element $a+J\left(R\right)$ and the set $\left\{\overline{a}\right|a\in A\}$, respectively. Then, from ([22] Lemma 2.1), we can see that $a\in R^{-1}$ if and only if $\overline{a}\in {\overline{R}}^{-1}$. Given $a\in R$, it is clear that $\overline{aR}=\overline{a}\overline{R}$. If R is a ∗-ring R, then $a\in J\left(R\right)$ if and only if $a^{\ast }\in J\left(R\right)$.

3. Characterizations for the Invertibility of ${\mathit{aa}}^{\parallel \mathit{d}}-{\mathit{a}}^{\parallel \mathit{d}}\mathit{a}$

The theme of this section is to focus on the equivalent conditions for the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$ under different conditions in rings, which extend the properties of co-EP elements. In particular, we find that this invertibility is related to the invertibility of the ring element expressed by certain functions of $a,d$ and suitable elements from the center of the ring. In addition, some examples are given to illustrate our results.

Now, we present the main result of this section as follows.

Theorem 1.

Let $a,d\in R$ be such that $a\in R^{\parallel d}$. Then, the following statements are equivalent:

(i)

$a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(ii)

$adR\oplus dR=R$ and $Rda\oplus Rd=R$.

(iii)

$adR+dR=R$, $Rda+Rd=R$, ${}^0\left(ad\right)$+${}^{0}d=R$ and ${\left(da\right)}^0+d^0=R$.

(iv)

$r={\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}a{d}^{2}a+{\lambda }_{4}d{a}^{2}d\in R^{-1}$ and one of the following conditions holds:

$\left(1\right)$

$adR\cap dR\subseteq J\left(R\right)$ and $Rda\cap Rd\subseteq J\left(R\right)$,

$\left(2\right)$

There exist $h,k\in R$ such that $had=ad$, $hd\in J\left(R\right)$, $dk=d$ and $dak\in J\left(R\right)$,

$\left(3\right)$

$d={\lambda }_{1}d{r}^{-1}ad={\lambda }_{2}da{r}^{-1}d$, ${\lambda }_1{\lambda }_{2}da{r}^{-1}ad=-{\lambda }_{4}d{a}^{2}d$ and ${\lambda }_1{\lambda }_{2}d{r}^{-1}d=-{\lambda }_3{d}^2$,

$\left(4\right)$

$1-a{\left(a^{\parallel d}\right)}^{2}a\in R^{-1}$,

where ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\lambda }_1{\lambda }_2\in R^{-1}$ and ${\lambda }_3{\lambda }_4\in a^0$.

(v)

$r={\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}a{d}^{2}a+{\lambda }_{4}d{a}^{2}d\in R^{-}$ and

$$adR\cap dR=\left\{0\right\},Rda\cap Rd=\left\{0\right\},{}^0\left(ad\right)\cap {}^{0}d=\left\{0\right\},{\left(da\right)}^0\cap d^0=\left\{0\right\}.$$

where ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\lambda }_1{\lambda }_2\in R^{-1}$ and ${\lambda }_3{\lambda }_4\in a^0$.

Proof. 

(i) ⇔ (ii). Applying the definition of the inverse along an element, we have

$$a{a}^{\parallel d}R=adR, a^{\parallel d}aR=dR, Ra{a}^{\parallel d}=Rd \mathrm{and} R{a}^{\parallel d}a=Rda.$$

Then, in view of Lemma 4 (i) and (ii), we deduce the equivalence of (i) and (ii).

(i) ⇒ (iii). By the equivalence (i) ⇔ (ii), we only need to show ${}^0\left(ad\right)$+${}^{0}d=R$ and ${\left(da\right)}^0+d^0=R$. Lemma 4 (i) and (v) ensures ${\left(a{a}^{\parallel d}\right)}^0+{\left(a^{\parallel d}a\right)}^0=R$. Note that ${\left(a^{\parallel d}a\right)}^0$=${\left(da\right)}^0$ and ${\left(a{a}^{\parallel d}\right)}^0$=$d^0$, which gives ${\left(da\right)}^0+d^0=R$. The proof of ${}^0\left(ad\right)$+${}^{0}d=R$ goes similarly.

(iii) ⇒ (ii). Since ${}^0\left(ad\right)$+${}^{0}d=R$, we obtain $1=p+q$, where $p\in$${}^0\left(ad\right)$ and $q\in$${}^{0}d$. Setting $s\in adR\cap dR$. Then, $s=ad{t}_1=d{t}_2$, for $t_1,t_2\in R$. Therefore, $s=1s=(p+q)s=pad{t}_1+qd{t}_2=0$, which implies $adR\cap dR=\left\{0\right\}$, so $adR\oplus dR=R$. Similarly, $Rda\oplus Rd=R$.

(i) ⇒ (iv). Suppose that ${\mathit{aa}}^{\parallel d}-a^{\parallel d}a\in R^{-1}$. Because of Lemma 5, there exist idempotents $h,k\in R$ satisfying (∗). Now, let

$$r^{\prime }={\lambda }_1(1-k){\left(da\right)}^{\#}(1-h)+{\lambda }_{2}k{\left(ad\right)}^{\#}h-{\lambda }_{3}k(1-h)-{\lambda }_4(1-k)h.$$

Observe that $a^{\parallel d}=d{\left(ad\right)}^{\#}={\left(da\right)}^{\#}d$ and ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\lambda }_3{\lambda }_4\in a^0$. Combining all these, we deduce that

$$\begin{array}{ccc}\hfill r{r}^{\prime }& \hfill =\hfill & \left({\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}a{d}^{2}a+{\lambda }_{4}d{a}^{2}d\right)·\hfill \\ & & \left({\lambda }_1(1-k){\left(da\right)}^{\#}(1-h)+{\lambda }_{2}k{\left(ad\right)}^{\#}h-{\lambda }_{3}k(1-h)-{\lambda }_4(1-k)h\right)\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_{2}ad{\left(ad\right)}^{\#}h-{\lambda }_1{\lambda }_{3}ad(1-h)+{\lambda }_1{\lambda }_{2}da{\left(da\right)}^{\#}(1-h)-{\lambda }_2{\lambda }_{4}dah\hfill \\ & & +{\lambda }_1{\lambda }_{3}a{d}^{2}a{\left(da\right)}^{\#}(1-h)+{\lambda }_2{\lambda }_{4}d{a}^{2}d{\left(ad\right)}^{\#}h\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_{2}a{a}^{\parallel d}h-{\lambda }_1{\lambda }_{3}ad(1-h)+{\lambda }_1{\lambda }_2{a}^{\parallel d}a(1-h)-{\lambda }_2{\lambda }_{4}dah\hfill \\ & & +{\lambda }_1{\lambda }_{3}ad{a}^{\parallel d}a(1-h)+{\lambda }_2{\lambda }_{4}d{a}^2{a}^{\parallel d}h\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_2\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill r^{\prime }r& \hfill =\hfill & \left({\lambda }_1(1-k){\left(da\right)}^{\#}(1-h)+{\lambda }_{2}k{\left(ad\right)}^{\#}h-{\lambda }_{3}k(1-h)-{\lambda }_4(1-k)h\right)·\hfill \\ & & \left({\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}a{d}^{2}a+{\lambda }_{4}d{a}^{2}d\right)\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_2(1-k){\left(da\right)}^{\#}da+{\lambda }_1{\lambda }_4(1-k){\left(da\right)}^{\#}d{a}^{2}d+{\lambda }_1{\lambda }_{2}k{\left(ad\right)}^{\#}ad\hfill \\ & & +{\lambda }_2{\lambda }_{3}k{\left(ad\right)}^{\#}a{d}^{2}a-{\lambda }_3{\lambda }_{2}kda-{\lambda }_1{\lambda }_4(1-k)ad\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_2(1-k)a^{\parallel d}a+{\lambda }_1{\lambda }_4(1-k)a^{\parallel d}{a}^{2}d+{\lambda }_1{\lambda }_{2}ka{a}^{\parallel d}\hfill \\ & & +{\lambda }_2{\lambda }_{3}ka{a}^{\parallel d}da-{\lambda }_3{\lambda }_{2}kda-{\lambda }_1{\lambda }_4(1-k)ad\hfill \\ & \hfill =\hfill & {\lambda }_1{\lambda }_2.\hfill \end{array}$$

Recall that ${\lambda }_1{\lambda }_2\in R^{-1}$. Thus, $r\in R^{-1}$ with $r^{-1}={\left({\lambda }_1{\lambda }_2\right)}^{-1}{r}^{\prime }$. By using the expression of $r^{-1}$, it is not difficult to check that item (3) holds. By reason of the equivalence (i) ⇔ (ii), we conclude that item (1) is true. In addition, it follows from Lemma 4 (i) and (iv) that $1-a{\left(a^{\parallel d}\right)}^{2}a\in R^{-1}$.

(iv) ⇒ (i) Firstly, suppose that $r\in R^{-1}$ and item (1) holds. Let $f=a{a}^{\parallel d}$ and $g=a^{\parallel d}a$. In order to prove $f-g\in R^{-1}$, it suffices to show $\overline{f}-\overline{g}\in {\overline{R}}^{-1}$. Note that $\overline{r}\in {\overline{R}}^{-1}$. So, we claim

$$\overline{1}=\overline{r}·{\overline{r}}^{-1}=\overline{ad}\left(\overline{{\lambda }_1}+\overline{{\lambda }_{3}da}\right){\overline{r}}^{-1}+\overline{d}\left(\overline{{\lambda }_{2}a}+\overline{{\lambda }_4{a}^{2}d}\right){\overline{r}}^{-1},$$

which gives that $\overline{ad}\overline{R}+\overline{d}\overline{R}=\overline{R}$. Let $\overline{x}\in \overline{ad}\overline{R}\cap \overline{d}\overline{R}$. Then, $\overline{x}=\overline{adu}=\overline{dv}$ for suitable $u,v\in R$. So, $j:=adu-dv\in J\left(R\right)$, which implies

$$adu-dv=1·j=r{r}^{-1}j=ad({\lambda }_1+{\lambda }_{3}da)r^{-1}j+d({\lambda }_{2}a+{\lambda }_4{a}^{2}d)r^{-1}j.$$

Thus,

$$j^{\prime }:=ad\left(u-({\lambda }_1+{\lambda }_{3}da)r^{-1}j\right)=d(v+({\lambda }_{2}a+{\lambda }_4{a}^{2}d)r^{-1}j)\in adR\cap dR\subseteq J\left(R\right).$$

Hence,

$$adu=j^{\prime }+ad({\lambda }_1+{\lambda }_{3}da)r^{-1}j\in J\left(R\right),$$

which leads to $\overline{x}=\overline{adu}=\overline{0}$. Hence, we deduce $\overline{ad}\overline{R}\cap \overline{d}\overline{R}=\left\{\overline{0}\right\}$, which implies $\overline{ad}\overline{R}\oplus \overline{d}\overline{R}=\overline{R}$. Note that $adR=fR$ and $dR=gR$. Therefore, $\overline{f}\overline{R}\oplus \overline{gR}=\overline{R}$. A similar argument yields $\overline{R}\overline{f}\oplus \overline{Rg}=\overline{R}$. Applying Lemma 4 (i) and (ii), we obtain $\overline{f}-\overline{g}\in {\overline{R}}^{-1}$, as required.

Secondly, suppose that $r\in R^{-1}$ and item (2) holds. By what we saw above, the condition $r\in R^{-1}$ implies $\overline{adR}+\overline{dR}=\overline{R}$ and $\overline{Rda}+\overline{Rd}=\overline{R}$. So, it only remains to verify that $\overline{adR}\cap \overline{dR}=\left\{\overline{0}\right\}$ and $\overline{Rda}\cap \overline{Rd}=\left\{\overline{0}\right\}$. Let $\overline{y}\in \overline{ad}\overline{R}\cap \overline{d}\overline{R}$. So, $\overline{y}=\overline{ad{h}_1}=\overline{d{h}_2}$, where $h_1,h_2\in R$. Then, applying item (2), we obtain $\overline{y}=\overline{had{h}_1}=\overline{hd{h}_2}=\overline{0}$, which gives $\overline{ad}\overline{R}\cap \overline{dR}=\left\{\overline{0}\right\}$. Similarly, $\overline{Rda}\cap \overline{Rd}=\left\{\overline{0}\right\}$.

Thirdly, suppose that $r\in R^{-1}$ and item (3) holds. Let

$$h^{\prime }=({\lambda }_{1}ad+{\lambda }_{3}a{d}^{2}a)r^{-1} \mathrm{and} k^{\prime }=r^{-1}({\lambda }_{1}ad+{\lambda }_{4}d{a}^{2}d).$$

By the hypotheses, we obtain

$$h^{\prime }ad=({\lambda }_{1}ad+{\lambda }_{3}a{d}^{2}a)r^{-1}ad={\lambda }_{1}ad{r}^{-1}ad-{\left({\lambda }_1{\lambda }_2\right)}^{-1}{\lambda }_3{\lambda }_{4}a{d}^2{a}^{2}d=ad$$

and

$$\begin{array}{ccc}\hfill h^{\prime }d& \hfill =\hfill & \left({\lambda }_{1}ad+{\lambda }_{3}a{d}^{2}a\right)r^{-1}d=\left(r-{\lambda }_{2}da-{\lambda }_{4}d{a}^{2}d\right)r^{-1}d\hfill \\ & \hfill =\hfill & d-{\lambda }_{2}da{r}^{-1}d-{\lambda }_{4}d{a}^{2}d{r}^{-1}d\hfill \\ & \hfill =\hfill & {\left({\lambda }_1{\lambda }_2\right)}^{-1}{\lambda }_3{\lambda }_{4}d{a}^2{d}^2\hfill \\ & \hfill =\hfill & 0\in J\left(R\right).\hfill \end{array}$$

Similarly, we have $d{k}^{\prime }=d$ and $da{k}^{\prime }\in J\left(R\right)$, which means that $h^{\prime },k^{\prime }$ satisfy item (2). This proves that item (i) holds.

Fourthly, suppose that $r\in R^{-1}$ and item (4) holds. By the invertibility of r, we conclude that $adR+dR=R$. If $z\in adR\cap dR$, then we get $z=ad{w}_1=d{w}_2$, for $w_1,w_2\in R$. Thus, we have

$$\begin{array}{ccc}\hfill \left(1-a{\left(a^{\parallel d}\right)}^{2}a\right)z& \hfill =\hfill & z-a{\left(a^{\parallel d}\right)}^{2}az=z-a{\left(a^{\parallel d}\right)}^{2}ad{w}_2=z-a{a}^{\parallel d}d{w}_2=z-a{a}^{\parallel d}z\hfill \\ & \hfill =\hfill & z-a{a}^{\parallel d}ad{w}_1=z-ad{w}_1=0.\hfill \end{array}$$

Applying $1-a{\left(a^{\parallel d}\right)}^{2}a\in R^{-1}$, we obtain $z=0$ and so $adR\cap dR=\left\{0\right\}$. Therefore, $adR\oplus dR=R$. Moreover, $Rda\oplus Rd=R$ can be obtained similarly. The statement $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$ now follows immediately from the equivalence (i) ⇔ (ii).

(i) ⇒ (v). By the equivalences of (i), (ii) and (iv), it therefore suffices to verify ${}^0\left(ad\right)$∩${}^{0}d=\left\{0\right\}$ and $d^0\cap {\left(da\right)}^0=\left\{0\right\}$. Let $m\in$${}^0\left(ad\right)$∩${}^{0}d$, keeping in mind that $adR+dR=R$. So, $ad{l}_1+d{l}_2=1$ for $l_1,l_2\in R$. Then, $m=m1=mad{l}_1+md{l}_2=0$, which gives that ${}^0\left(ad\right)$∩${}^{0}d=\left\{0\right\}$. Similarly, we obtain ${\left(da\right)}^0\cap d^0=\left\{0\right\}$.

(v) ⇒ (i). First, we show that $r^0=\left\{0\right\}$. For any $n\in r^0$, we have

$$0=rn={\lambda }_{1}adn+{\lambda }_{2}dan+{\lambda }_{3}a{d}^{2}an+{\lambda }_{4}d{a}^{2}dn,$$

which implies

$${\lambda }_{2}dan+{\lambda }_{4}d{a}^{2}dn=ad(-{\lambda }_{1}n-{\lambda }_{3}dan)\in adR\cap dR=\left\{0\right\}.$$

So, ${\lambda }_{2}dan+{\lambda }_{4}d{a}^{2}dn=0=-{\lambda }_{1}adn-{\lambda }_{3}a{d}^{2}an$. Multiplying the equality ${\lambda }_{2}dan+{\lambda }_{4}d{a}^{2}dn=0$ by ${\lambda }_3$ from the left, we obtain ${\lambda }_2{\lambda }_{3}dan=0$, i.e., ${\lambda }_{3}dan=0$. Thus, ${\lambda }_{1}adn=-{\lambda }_{3}a{d}^{2}an=0$, i.e., $adn=0$. Consequently, $dn=a^{\parallel d}adn=0$ and $dan=-{\lambda }_2^{-1}{\lambda }_{4}d{a}^{2}dn=0$. Therefore, $n\in$$d^0\cap {\left(da\right)}^0=\left\{0\right\}$, which means $n=0$; then, $r^0=\left\{0\right\}$. Since $r\in R^{-}$, there exists $r^{-}\in R$ such that $r(1-r^{-}r)=0$, Hence, we conclude $r^{-}r=1$. Analogously, $r{r}^{-}=1$, and thus $r\in R^{-1}$. Applying the implication (iv) ⇒ (i), which has been proved before, we claim that item (i) holds. □

Remark 1.

(a) The following example shows that the implication (i) ⇒ (iv) in Theorem 1 does not hold in general if one of the assumptions, ${\lambda }_1{\lambda }_2\in R^{-1}$ or ${\lambda }_3{\lambda }_4\in a^0$, is dropped. For example, setting $R={\mathbb{C}}^{2\times 2}$ and choosing $a=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, $d=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\in R$. Then, we deduce that $a^{\parallel d}=d$ and $a{a}^{\parallel d}-a^{\parallel d}a=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\in R^{-1}$. $\left(1\right)$ Let ${\lambda }_1={\lambda }_2=I,{\lambda }_3=-Iand{\lambda }_4=-\frac{1}{2}I$, where I is the identity matrix. Obviously, ${\lambda }_1{\lambda }_2\in R^{-1}$. However, we have $r=\left(\begin{array}{cc}0& 0\\ 0& -1\end{array}\right)\notin R^{-1}$. $\left(2\right)$ Let ${\lambda }_1={\lambda }_2={\lambda }_3={\lambda }_4=0$. Clearly, ${\lambda }_3{\lambda }_4\in a^0$. However, $r=0\notin R^{-1}$.

$\left(b\right)$ When $a\in R^{\parallel d}$, the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$ can also imply the invertibility of s; here,

$$s={\mu }_{1}ad+{\mu }_2{a}^{\parallel d}a+{\mu }_{3}ad{a}^{\parallel d}a+{\mu }_4{a}^{\parallel d}{a}^{2}d,$$

where ${\mu }_i\in C\left(R\right) (i\in \overline{1,4})$, ${\mu }_1{\mu }_2\in R^{-1}$ and ${\mu }_3{\mu }_4\in a^0$, as well as the invertibility of t. Here

$$t={\xi }_{1}a{a}^{\parallel d}+{\xi }_2{a}^{\parallel d}a+{\xi }_{3}a{\left(a^{\parallel d}\right)}^{2}a+{\xi }_4{a}^{\parallel d}{a}^2{a}^{\parallel d},$$

where ${\xi }_i\in C\left(R\right)(i\in \overline{1,4})$, ${\xi }_1{\xi }_2\in R^{-1}$ and ${\xi }_3{\xi }_4\in a^0$. Indeed, let $h,k$ be defined as in Lemma 5, then we can directly check that

$$s^{-1}={\left({\mu }_1{\mu }_2\right)}^{-1}\left({\mu }_1(1-k)(1-h)+{\mu }_{2}k{\left(ad\right)}^{\#}h-{\mu }_{3}k(1-h)-{\mu }_4(1-k)h\right)$$

and

$$t^{-1}={\left({\xi }_1{\xi }_2\right)}^{-1}\left({\xi }_1(1-k)(1-h)+{\xi }_{2}kh-{\xi }_{3}k(1-h)-{\xi }_4(1-k)h\right).$$

If we replace the condition $a\in R^{\parallel d}$ with $a\in R^{\parallel •d}$ in Theorem 1, then we further obtain the following result, which can be proved by using the same strategy as the proof of Theorem 1. Here, we give a simple proof for it.

Theorem 2.

Let $a,d\in R$ be such that $a\in R^{\parallel •d}$. Then, the following statements are equivalent:

(i)

$a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(ii)

$d{d}^{\parallel a}-d^{\parallel a}d\in R^{-1}$.

(iii)

$aR\oplus dR=R$ and $Ra\oplus Rd=R$.

(iv)

$aR+dR=R$, $Ra+Rd=R$, ${}^{0}a$+${}^{0}d=R$ and $a^0+d^0=R$.

(v)

One of the following conditions holds:

(1)

$m={\delta }_{1}a+{\delta }_{2}d+{\delta }_{3}ad+{\delta }_{4}da\in R^{-1}$, where ${\delta }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\delta }_1{\delta }_2\in R^{-1}$, ${\delta }_3{\delta }_4\in a^0$;

(2)

$n={\gamma }_{1}a+{\gamma }_{2}ad+{\gamma }_{3}da+{\gamma }_{4}a{d}^{2}a\in R^{-1}$, where ${\gamma }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\gamma }_2{\gamma }_3\in R^{-1}$;

and one of the following conditions holds:

(1′)

$aR\cap dR\subseteq J\left(R\right)$ and $Ra\cap Rd\subseteq J\left(R\right)$;

(2′)

There exist $h,k\in R$ such that $ha=a$, $hd\in J\left(R\right)$, $dk=d$ and $ak\in J\left(R\right)$.

(vi)

One of the following conditions holds:

$\left(1\right)$

$m={\delta }_{1}a+{\delta }_{2}d+{\delta }_{3}ad+{\delta }_{4}da\in R^{-}$, where ${\delta }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\delta }_1{\delta }_2\in R^{-1}$, ${\delta }_3{\delta }_4\in a^0$;

$\left(2\right)$

$n={\gamma }_{1}a+{\gamma }_{2}ad+{\gamma }_{3}da+{\gamma }_{4}a{d}^{2}a\in R^{-}$, where ${\gamma }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\gamma }_2{\gamma }_3\in R^{-1}$,

and the following condition holds:

$$aR\cap dR=\left\{0\right\}, Ra\cap Rd=\left\{0\right\}, {}^{0}a\cap {}^{0}d=\left\{0\right\}, a^0\cap d^0=\left\{0\right\}.$$

Proof. 

(i) ⇔ (ii) follows directly by Lemma 3.

(i) ⇔ (iii) ⇔ (iv). Since $a\in R^{\parallel •d}$, we have $a=a{a}^{\parallel d}a=ad{\left(ad\right)}^{\#}a=ad{\left({\left(ad\right)}^{\#}\right)}^{2}ada$. Hence, $adR=aR$, $Rda=Ra$, ${}^0\left(ad\right)$=${}^{0}a$ and ${\left(da\right)}^0=a^0$, which leads to the equivalences of (i), (iii) and (iv) by Theorem 1 (i)–(iii).

(i) ⇒ (v). As in the proof of Theorem 1, by $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$, it follows that there exist idempotents $h,k\in R$ satisfying (∗). Moreover,

$$ha=ha{a}^{\parallel d}a=\left(had\right){\left(ad\right)}^{\#}a=ad{\left(ad\right)}^{\#}a=a{a}^{\parallel d}a=a$$

and

$$ak=a{a}^{\parallel d}ak=a{\left(da\right)}^{\#}\left(dak\right)=0.$$

Hence, item (2′) holds. Note that $adR=aR$ and $Rda=Ra$. Then, by Theorem 1 (i) and (iv), we obtain that item (1′) is true.

Next, by a direct computation, we conclude that $m,n\in R^{-1}$ and

$$m^{-1}={\left({\delta }_1{\delta }_2\right)}^{-1}\left({\delta }_{1}ka{\left(da\right)}^{\#}(1-h)+{\delta }_2(1-k)a^{\parallel d}h-{\delta }_3(1-k)(1-h)-{\delta }_{4}kh\right),$$

$$n^{-1}={\left({\gamma }_2{\gamma }_3\right)}^{-1}\left(-{\gamma }_{1}k{d}^{\parallel a}{a}^{\parallel d}{d}^{\parallel a}(1-h)+{\gamma }_2(1-k){\left(da\right)}^{\#}(1-h)+{\gamma }_{3}k{\left(ad\right)}^{\#}h-{\gamma }_{4}k(1-h)\right).$$

(v) ⇒ (i). Assume that item (v) holds. Setting $f=a{a}^{\parallel d}$ and $g=a^{\parallel d}a$. Similar to the proof of Theorem 1 (iv) ⇒ (i), we obtain $\overline{aR}\oplus \overline{dR}=\overline{R}$. Note that $aR=fR$ and $dR=gR$. Hence, $\overline{f}\overline{R}\oplus \overline{gR}=\overline{R}$. On the other hand, $\overline{R}\overline{f}\oplus \overline{Rg}=\overline{R}$. So, $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(i) ⇔ (vi). We check this part in a manner analogous to Theorem 1 (i) ⇔ (v). □

In Theorem 2, the condition $a\in R^{\parallel •d}$ cannot be replaced by $a\in R^{\parallel d}$ in general. The following example illustrates that under the condition $a\in R^{\parallel d}$, item (i) holds, but item (v) fails to hold.

Example 1.

Let $R={\mathbb{C}}^{2\times 2}$. Choose $a=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)andd=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).$ Then, we can check that $a^{\parallel d}=d$ and $a{a}^{\parallel d}-a^{\parallel d}a=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\in R^{-1}$. However, $a+d=a+a^{\parallel d}\notin R^{-1}$ and $a+ad-da\notin R^{-1}$.

Let us mention that $a\in R^{†}$ if and only if $a\in R^{\parallel a^{\ast }}$, if and only if $a\in R^{\parallel •a^{\ast }}$. Thus, on the basis of Theorems 1 and 2, we obtain some new results about the characterizations of co-EP elements as follows.

Corollary 1.

Let R be a ∗-ring and $a\in R^{†}$. Then, the following statements are equivalent:

(i)

$a\in R_{co}^{\mathrm{EP}}$.

(ii)

$aR$+$a^{\ast }R=R$ and ${}^{0}a$+${}^0\left(a^{\ast }\right)=R$.

(iii)

One of the following conditions holds:

$\left(1\right)$

${\lambda }_{1}a{a}^{\ast }+{\lambda }_2{a}^{\ast }a+{\lambda }_{3}a{\left(a^{\ast }\right)}^{2}a+{\lambda }_4{a}^{\ast }{a}^2{a}^{\ast }\in R^{-1}$, where ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\lambda }_1{\lambda }_2\in R^{-1}$, ${\lambda }_3{\lambda }_4\in a^0$;

$\left(2\right)$

${\delta }_{1}a+{\delta }_2{a}^{\ast }+{\delta }_{3}a{a}^{\ast }+{\delta }_4{a}^{\ast }a\in R^{-1}$, where ${\delta }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\delta }_1{\delta }_2\in R^{-1}$, ${\delta }_3{\delta }_4\in a^0$;

$\left(3\right)$

${\gamma }_{1}a+{\gamma }_{2}a{a}^{\ast }+{\gamma }_3{a}^{\ast }a+{\gamma }_{4}a{\left(a^{\ast }\right)}^{2}a\in R^{-1}$, where ${\gamma }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\gamma }_2{\gamma }_3\in R^{-1}$,

and one of the following conditions holds:

$\left(1^{\prime }\right)$

$aR\cap a^{\ast }R\subseteq J\left(R\right)$;

$\left(2^{\prime }\right)$

There exists $h\in R$ such that $ha=a$ and $h{a}^{\ast }\in J\left(R\right)$.

(iv)

One of the following conditions holds:

$\left(1\right)$

${\lambda }_{1}a{a}^{\ast }+{\lambda }_2{a}^{\ast }a+{\lambda }_{3}a{\left(a^{\ast }\right)}^{2}a+{\lambda }_4{a}^{\ast }{a}^2{a}^{\ast }\in R^{-}$, where ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\lambda }_1{\lambda }_2\in R^{-1}$, ${\lambda }_3{\lambda }_4\in a^0$;

$\left(2\right)$

${\delta }_{1}a+{\delta }_2{a}^{\ast }+{\delta }_{3}a{a}^{\ast }+{\delta }_4{a}^{\ast }a\in R^{-}$, where ${\delta }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\delta }_1{\delta }_2\in R^{-1}$, ${\delta }_3{\delta }_4\in a^0$;

$\left(3\right)$

${\gamma }_{1}a+{\gamma }_{2}a{a}^{\ast }+{\gamma }_3{a}^{\ast }a+{\gamma }_{4}a{\left(a^{\ast }\right)}^{2}a\in R^{-}$, where ${\gamma }_i\in C\left(R\right)$$(i\in \overline{1,4})$, ${\gamma }_2{\gamma }_3\in R^{-1}$,

and $aR\cap a^{\ast }R=\left\{0\right\}$, ${}^{0}a$∩${}^0\left(a^{\ast }\right)=\left\{0\right\}.$

Proof. 

Taking $d=a^{\ast }$ in Theorems 1 and 2, we deduce that Corollary 1 holds by the following facts:

• $aR=a{a}^{\ast }R$ and $Ra=R{a}^{\ast }a$ because of $a\in R^{†}$ ([2] Theorem 5.4).
• ${}^0\left(a{a}^{\ast }\right)$=${}^{0}a$ and ${\left(a^{\ast }a\right)}^0$=$a^0$, because a is ∗-cancelable.
• The next equivalences are obvious by taking the involution:

  (a)

  ${}^{0}a+$${}^0{a}^{\ast }=R$$\iff a^0+{\left(a^{\ast }\right)}^0=R$.

  (b)

  ${}^{0}a$${\cap }^0\left(a^{\ast }\right)=\left\{0\right\}$$\iff a^0\cap {\left(a^{\ast }\right)}^0=\left\{0\right\}$.

  (c)

  $aR+a^{\ast }R=R\iff Ra+R{a}^{\ast }=R$.

  (d)

  $aR\cap a^{\ast }R=\left\{0\right\}\iff Ra\cap R{a}^{\ast }=\left\{0\right\}$.

  (e)

  $aR\cap a^{\ast }R\subseteq J\left(R\right)\iff Ra\cap R{a}^{\ast }\subseteq J\left(R\right)$.

• Let $h\in R$ be such that $ha=a$ and $h{a}^{\ast }\in J\left(R\right)$. Then, the element $k:=h^{\ast }$ satisfies $a^{\ast }k=a^{\ast }$ and $a^{\ast }ak\in J\left(R\right)$, owing to ($a^{\ast }a{h}^{\ast }\in J\left(R\right)\iff h{a}^{\ast }a\in J\left(R\right)\iff h{a}^{\ast }=h{a}^{\ast }a{a}^{†}\in J\left(R\right)$) and ($a^{\ast }{h}^{\ast }=a^{\ast }\iff ha=a$).

□

In what follows, we further discuss some equivalent conditions for the invertibility of $a{a}^{\parallel d}-a^{\parallel d}a$, when both a and d are idempotents. By $R^{\mathrm{idem}}$, we denote the set of all idempotents in R.

Theorem 3.

Let $a,d\in R^{\mathrm{idem}}$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$ and $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(ii)

$d\in R^{\parallel a}$ and $d{d}^{\parallel a}-d^{\parallel a}d\in R^{-1}$.

(iii)

For ${\lambda }_i\in C\left(R\right)$$(i\in \overline{1,4})$ such that ${\lambda }_1{\lambda }_2\in R^{-1}$, ${\lambda }_3{\lambda }_4\in a^0$ and ${\displaystyle \sum _{i=1}^4}{\lambda }_i=0$, we have

$r={\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}ada+{\lambda }_{4}dad\in R^{-1}$.

(iv)

For ${\delta }_i\in C\left(R\right)$$(i\in \overline{1,4})$ such that ${\delta }_1{\delta }_2\in R^{-1}$, ${\delta }_3{\delta }_4\in a^0$ and ${\displaystyle \sum _{i=1}^4}{\delta }_i=0$, we have

$a\in R^{\parallel •d}$ and $m={\delta }_{1}a+{\delta }_{2}d+{\delta }_{3}ad+{\delta }_{4}da\in R^{-1}$.

(v)

For ${\gamma }_i\in C\left(R\right)$$(i\in \overline{1,4})$ such that ${\gamma }_2{\gamma }_3\in R^{-1}$ and ${\displaystyle \sum _{i=1}^4}{\gamma }_i=0$, we have

$a\in R^{\parallel •d}$ and $n={\gamma }_{1}a+{\gamma }_{2}ad+{\gamma }_{3}da+{\gamma }_{4}ada\in R^{-1}$.

Proof. 

(i) ⇒ (iii) follows directly from Theorem 1.

(iii) ⇒ (i). According to the hypotheses, we have

$$a=ar{r}^{-1}=({\lambda }_{1}ad+{\lambda }_{2}ada+{\lambda }_{3}ada+{\lambda }_{4}adad)r^{-1}\in adR$$

and

$$d=dr{r}^{-1}=({\lambda }_{1}dad+{\lambda }_{2}da+{\lambda }_{3}dada+{\lambda }_{4}dad)r^{-1}\in daR,$$

which implies that $d\in dadR$. Similarly, $d\in Rdad$. Then, Lemma 2 quickly leads to $a\in R^{\parallel d}$.

By Theorem 1 (i) and (iv), it remains to show $adR\cap dR\subseteq J\left(R\right)$ and $Rda\cap Rd\subseteq J\left(R\right)$. Let $x\in adR\cap dR$. So, $x=adu=dv$ for $u,v\in R$. Since $a,d\in R^{\mathrm{idem}}$, we have $ax=aadu=adu=x=dv=d\left(dv\right)=dx$, which gives that

$$x=r^{-1}\left({\lambda }_{1}ad+{\lambda }_{2}da+{\lambda }_{3}ada+{\lambda }_{4}dad\right)x=r^{-1}\sum _{i=1}^4{\lambda }_{i}x=0.$$

Thus, $adR\cap dR=\left\{0\right\}\subseteq J\left(R\right)$. Similarly, $Rda\cap Rd\subseteq J\left(R\right)$, as required.

(ii) ⇔ (iii) can be obtained by the equivalence (i) ⇔ (iii) and the symmetry of $a,d$.

(i) ⇔ (iv). Suppose that item (i) holds. So, using the equivalence (i) ⇔ (ii), we obtain $d\in R^{\parallel a}$. This proves $a\in R^{\parallel •d}$ by Lemma 3. We can now make use of Theorem 2 (i) and (v) to show that item (iv) holds.

Conversely, as in the proof of the implication (iii) ⇒ (i), we deduce that $aR\cap dR\subseteq J\left(R\right)$ and $Ra\cap Rd\subseteq J\left(R\right)$. Applying Theorem 2 (i) and (v) again, we obtain $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(i) ⇔ (v) can be proved following similar steps to those in the proof of the equivalence (i) ⇔ (iv). □

In the light of Theorem 3, we derive

Corollary 2.

Let $a,d\in R^{\mathrm{idem}}$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$ and $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$.

(ii)

$d\in R^{\parallel a}$ and $d{d}^{\parallel a}-d^{\parallel a}d\in R^{-1}$.

(iii)

$ad-da\in R^{-1}$.

(iv)

$a\in R^{\parallel •d}$ and $a-d\in R^{-1}$.

(v)

$a\in R^{\parallel •d}$ and $2a-ad-da\in R^{-1}$.

In general, the idempotence of d in Corollary 2 cannot be dropped, as is shown by the next example.

Example 2.

In the ring $R=M_2\left({\mathbb{Z}}_4\right)$, let

$$a=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)andd=\left(\begin{array}{cc}2& 1\\ 1& 0\end{array}\right).$$

Clearly, $a=a^2$ and $ad-da\in R^{-1}$. Therefore, a and d satisfy item $\left(iii\right)$ of Corollary 2. However, $ad\notin R^{\#}$, which means $a\notin R^{\parallel d}$. Hence, item $\left(i\right)$ does not hold.

In the same way, the idempotence of a in Corollary 2 cannot be deleted, either. This can be seen from the following example.

Example 3.

In the ring $R=M_2\left({\mathbb{Z}}_2\right)$, let

$$a=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)andd=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right).$$

Obviously, $d=d^2$, $a^{\parallel d}=d$ and $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$. So, a and d satisfy item $\left(i\right)$ of Corollary 2. However, $d\notin R^{\parallel a}$. Thus, item $\left(ii\right)$ does not hold.

We close this section by considering two classes of special idempotents.

Remark 2.

Let R be a ring. For $x,y\in R$, consider the following idempotents in $M_2\left(R\right)$,

$$a=\left(\begin{array}{cc}1& x\\ 0& 0\end{array}\right)andd=\left(\begin{array}{cc}0& 0\\ y& 1\end{array}\right).$$

Then, we have the following fact:

$$a\in R^{\parallel d} \mathit{and} a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}⟺yx \in R^{-1} \mathit{and} x\left({\left(yx\right)}^{-1}-{\left(yx\right)}^{-2}\right)y\in R^{-1}.$$

Indeed, by Lemma 2, we obtain that $a\in R^{\parallel d}$ if and only if $yx\in R^{-1}$. In this case, we have

$$a^{\parallel d}=\left(\begin{array}{cc}0& 0\\ {\left(yx\right)}^{-1}y& {\left(yx\right)}^{-1}\end{array}\right)anda{a}^{\parallel d}-a^{\parallel d}a=\left(\begin{array}{cc}x{\left(yx\right)}^{-1}y& x{\left(yx\right)}^{-1}\\ -{\left(yx\right)}^{-1}y& -1\end{array}\right).$$

Clearly, $a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}$ if and only if $\left(x{\left(yx\right)}^{-1}y\right)·(-1)-\left(x{\left(yx\right)}^{-1}\right)·\left(-{\left(yx\right)}^{-1}y\right)\in R^{-1}$. Furthermore, if R is a commutative ring, then

$$a\in R^{\parallel d}\mathit{and}a{a}^{\parallel d}-a^{\parallel d}a\in R^{-1}⟺xy,1-xy\in R^{-1}⟺a\in R^{\parallel d}\mathit{and}a+d\in R^{-1}.$$

From Remark 1 (b), we see that the element d can be replaced by $a^{\parallel d}$, which motivates us to consider the same properties between d and $a^{\parallel d}$, as follows.

Theorem 4.

Let $a,d\in R$ be such that $a\in R^{\parallel d}$ (R is a ∗-ring in (ii)–(iv)). Then, we have the following:

(i)

$d\in R^{\#}$ if and only if $a^{\parallel d}\in R^{\#}$. In this case, ${\left(a^{\parallel d}\right)}^{\#}=d^{\#}dad{d}^{\#}$.

(ii)

$d\in R^{\{1,3\}}$ if and only if $a^{\parallel d}\in R^{\{1,3\}}$. In this case, $ad{d}^{(1,3)}$ is a $\{1,3\}$-inverse of $a^{\parallel d}$.

(iii)

$d\in R^{\{1,4\}}$ if and only if $a^{\parallel d}\in R^{\{1,4\}}$. In this case, $d^{(1,4)}da$ is a $\{1,4\}$-inverse of $a^{\parallel d}$.

(iv)

$d\in R^{†}$ if and only if $a^{\parallel d}\in R^{†}$. In this case, ${\left(a^{\parallel d}\right)}^{†}=d^{†}dad{d}^{†}$.

Proof. 

(i). Assume that $d\in R^{\#}$. Since $a\in R^{\parallel d}$, we obtain

$$a^{\parallel d}=d{\left(ad\right)}^{\#}=d^{\#}dd{\left(ad\right)}^{\#}=d^{\#}d{a}^{\parallel d}=d^{\#}da{\left(a^{\parallel d}\right)}^2\in R{\left(a^{\parallel d}\right)}^2.$$

Dually, $a^{\parallel d}={\left(a^{\parallel d}\right)}^{2}ad{d}^{\#}\in {\left(a^{\parallel d}\right)}^{2}R$. So, $a^{\parallel d}\in R^{\#}$ and ${\left(a^{\parallel d}\right)}^{\#}=d^{\#}dad{d}^{\#}$ by Lemma 7.

Conversely, if $a^{\parallel d}\in R^{\#}$, then

$$d=da{a}^{\parallel d}=da{a}^{\parallel d}{a}^{\parallel d}{\left(a^{\parallel d}\right)}^{\#}=d{a}^{\parallel d}{\left(a^{\parallel d}\right)}^{\#}=d^2{\left(ad\right)}^{\#}{\left(a^{\parallel d}\right)}^{\#}\in d^{2}R.$$

On the other hand, we can infer that $d\in R{d}^2$. Hence, $d\in R^{\#}$.

(ii). Assume that $d\in R^{\{1,3\}}$. Then, it is clear that

$$\begin{array}{ccc}\hfill a^{\parallel d}& \hfill =\hfill & d{\left(ad\right)}^{\#}=d{d}^{(1,3)}d{\left(ad\right)}^{\#}={\left(d{d}^{(1,3)}\right)}^{\ast }{a}^{\parallel d}={\left(d^{(1,3)}\right)}^{\ast }{d}^{\ast }{a}^{\parallel d}\hfill \\ & \hfill =\hfill & {\left(d^{(1,3)}\right)}^{\ast }{\left(a^{\parallel d}ad\right)}^{\ast }{a}^{\parallel d}={\left(d^{(1,3)}\right)}^{\ast }{d}^{\ast }{a}^{\ast }{\left(a^{\parallel d}\right)}^{\ast }{a}^{\parallel d}\hfill \\ & \hfill =\hfill & d{d}^{(1,3)}{a}^{\ast }\left({\left(a^{\parallel d}\right)}^{\ast }{a}^{\parallel d}\right).\hfill \end{array}$$

Hence, from Lemma 8 (i), it follows that $a^{\parallel d}\in R^{\{1,3\}}$ and $ad{d}^{(1,3)}$ is a $\{1,3\}$-inverse of $a^{\parallel d}$.

Now, let $a^{\parallel d}\in R^{\{1,3\}}$. Then,

$$\begin{array}{ccc}\hfill d& \hfill =\hfill & a^{\parallel d}ad=a^{\parallel d}{\left(a^{\parallel d}\right)}^{(1,3)}{a}^{\parallel d}ad={\left(a^{\parallel d}{\left(a^{\parallel d}\right)}^{(1,3)}\right)}^{\ast }d={\left({\left(a^{\parallel d}\right)}^{(1,3)}\right)}^{\ast }{\left(a^{\parallel d}\right)}^{\ast }d\hfill \\ & \hfill =\hfill & {\left({\left(a^{\parallel d}\right)}^{(1,3)}\right)}^{\ast }{\left(d{\left(ad\right)}^{\#}\right)}^{\ast }d={\left({\left(a^{\parallel d}\right)}^{(1,3)}\right)}^{\ast }{\left({\left(ad\right)}^{\#}\right)}^{\ast }{d}^{\ast }d\hfill \\ & \hfill =\hfill & {\left({\left(ad\right)}^{\#}{\left(a^{\parallel d}\right)}^{(1,3)}\right)}^{\ast }\left(d^{\ast }d\right).\hfill \end{array}$$

Hence, $d\in R^{\{1,3\}}$.

(iii). This can be obtained in the same way as the proof of item (ii).

(iv). Note that $a^{\parallel d}\in R^{†}$ if and only if $a^{\parallel d}\in R^{\{1,3\}}\cap R^{\{1,4\}}$. Then, applying items (ii) and (iii), together with ${\left(a^{\parallel d}\right)}^{†}={\left(a^{\parallel d}\right)}^{(1,4)}{a}^{\parallel d}{\left(a^{\parallel d}\right)}^{(1,3)}$, we claim that item (iv) holds. □

4. Characterizations for the Equality ${\mathit{aa}}^{\parallel \mathit{d}}={\mathit{a}}^{\parallel \mathit{d}}\mathit{a}$

Given $a,d\in R$, if $a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$, then $a^{\parallel d}$ is called the commutative inverse of a along d [23]. Benítez and Boasso took advantage of the group invertibility to characterize the commutative inverse along an element and obtained some equivalent conditions for this inverse under the condition $a\in R^{\parallel d}$. Recently, Wang, Mosić and Yao [24] also studied this topic. In this section, we continue to consider it under the new hypothesis $d\in R^{\parallel a}$.

We start this part with two useful lemmas on which we rely.

Lemma 9

([23] Theorems 7.1 and 7.3). Let $a,d\in R$ be such that $a\in R^{\parallel d}$. Then, the following statements are equivalent:

(i)

$a{a}^{\parallel d}=a^{\parallel d}a$.

(ii)

$d\in R^{\#}$ and $ad{d}^{\#}=d{d}^{\#}a$.

(iii)

$da\in Rd$ and $ad\in dR$.

Lemma 10.

Let $a,d\in R$ be such that $d\in R^{\parallel a}$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$.

(ii)

$aR=dR$ and $Ra=Rd$.

(iii)

$d\in R^{\#}$, $aR\subseteq dR$ and $Ra=Rd$.

(iv)

$a\in R^{\parallel d}$, $aR\subseteq dR$ and $Ra\subseteq Rd$.

Proof. 

By the condition $d\in R^{\parallel a}$, it follows that $a=ada{t}_1$ and $a=t_{2}ada$ for suitable $t_1,t_2\in R$. Moreover, we obtain $ad\in R^{\#}$ and $da\in R^{\#}$.

(i) ⇒ (ii). Suppose that $a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$. Then, by Lemma 3, we deduce $a=a^{\parallel d}{a}^2=d{d}^{\parallel a}a\in dR$ and $d=a^{\parallel d}ad=a{a}^{\parallel d}d\in aR$. Hence, $aR=dR$. By a symmetric argument, we have $Ra=Rd$.

(ii) ⇒ (iii). By $aR=dR$, we have $d=au$ for some $u\in R$. Then, multiplying the equality $a=t_{2}ada$ by u from the right side, we obtain $d=t_{2}a{d}^2\in R{d}^2$. Similarly, using $Ra=Rd$, we deduce that $d\in d^{2}R$. So, in view of Lemma 7, we obtain $d\in R^{\#}$.

(iii) ⇒ (iv). We only need to show that $a\in R^{\parallel d}$. Note that $d=va$, where $v\in R$. Hence,

$$d=d^{\#}dd=d^{\#}vad=d^{\#}vada{t}_{1}d=d^{\#}dda{t}_{1}d=da{t}_{1}d\in daR.$$

By Lemma 1 (i) and (ii), we obtain $a\in R^{\parallel d}$.

(iv) ⇒ (i). Applying the hypotheses $aR\subseteq dR$ and $Ra\subseteq Rd$, we directly obtain $da\in Rd$ and $ad\in dR$. In the light of Lemma 9 (i) and (iii), we claim that item (i) holds. □

Next, we present an existence criterion for the commutative inverse along an element, which covers the related EP properties.

Theorem 5.

Let $a,d\in R$ be such that $d\in R^{\parallel a}$ and $k\in \mathbb{N}$, $m\in \mathbb{N}\cup \left\{0\right\}$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$.

(ii)

$a,d\in R^{\#}$ and $a{a}^{\#}=d{d}^{\#}$.

(iii)

$a\in R^{\parallel d}\cap R^{\#}$ and ${\left(a^{\parallel d}\right)}^k={\left(a^{\#}\right)}^k$.

(iv)

$a\in R^{\parallel d}$, $d{\left(ad\right)}^m\in R^{\#}$ and ${\left(d{\left(ad\right)}^m\right)}^{\#}=d^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^m$.

Proof. 

(i) ⇒ (ii), (iii) and (iv). Suppose that $a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$. By Lemma 3, we have $d{d}^{\parallel a}=d^{\parallel a}d$. Lemma 9 now yields $a,d\in R^{\#}$, $da\in Rd$ and $ad\in dR$. So, $da=sd$, for $s\in R$. Then, $dad{d}^{\#}=sdd{d}^{\#}=sd=da$. Hence, $a^{\parallel d}a={\left(da\right)}^{\#}da={\left(da\right)}^{\#}dad{d}^{\#}=a^{\parallel d}ad{d}^{\#}=d{d}^{\#}$. Similarly, $d^{\parallel a}d=a{a}^{\#}$, i.e., $a{a}^{\parallel d}=a{a}^{\#}$. Therefore, $a{a}^{\#}=d{d}^{\#}$. Moreover, $a^{\parallel d}=a^{\parallel d}a{a}^{\parallel d}=a^{\parallel d}a{a}^{\#}=d^{\#}d{a}^{\#}=a^{\#}a{a}^{\#}=a^{\#}$, and correspondingly, $d^{\parallel a}=d^{\#}$. Let $x=d^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^m$. Then,

$$\begin{array}{ccc}\hfill d{\left(ad\right)}^{m}x& \hfill =\hfill & d{\left(ad\right)}^m{d}^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^m=d{\left(ad\right)}^m{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^m\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}a\left(d{d}^{\#}\right)a^{\#}{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}aa{a}^{\#}{a}^{\#}{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}a{a}^{\#}{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^{m-1}\hfill \\ & \hfill ⋮\hfill & \\ & \hfill =\hfill & d{d}^{\#}.\hfill \end{array}$$

In the same manner, we obtain $xd{\left(ad\right)}^m=d^{\#}d$, which gives $d{\left(ad\right)}^{m}x=xd{\left(ad\right)}^m$. Moreover, we conclude $xd{\left(ad\right)}^{m}x=d^{\#}d{d}^{\#}{\left(a^{\#}{d}^{\#}\right)}^m=d^{\#}{\left(a^{\#}{d}^{\#}\right)}^m=x$ and $d{\left(ad\right)}^{m}xd{\left(ad\right)}^m=d{d}^{\#}d{\left(ad\right)}^m=d{\left(ad\right)}^m$. So, $d{\left(ad\right)}^m\in R^{\#}$ and ${\left(d{\left(ad\right)}^m\right)}^{\#}=x$.

(ii) ⇒ (i). By the hypotheses, we obtain $a=a{a}^{\#}a=d{d}^{\#}a\in dR$ and $d=d{d}^{\#}d=a{a}^{\#}d\in aR$, which show that $aR=dR$. Similarly, we have $Ra=Rd$. Applying Lemma 10 (i) and (ii), we obtain $a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$.

(iii) ⇒ (i). It is clear that $a=a^{k+1}{\left(a^{\#}\right)}^k=a^{k+1}{\left(a^{\parallel d}\right)}^{k-1}{\left(da\right)}^{\#}d\in Rd$, which yields $Ra\subseteq Rd$. On the other hand, we have $aR\subseteq dR$. From Lemma 10 (i) and (iv), it follows that item (i) holds.

(iv) ⇒ (i). Note that $d^{\parallel a}da=a=ad{d}^{\parallel a}$ and $a^{\parallel d}ad=d=da{a}^{\parallel d}$. So, we have

$$\begin{array}{ccc}\hfill d{\left(ad\right)}^m{\left(d{\left(ad\right)}^m\right)}^{\#}& \hfill =\hfill & d{\left(ad\right)}^m{d}^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^m\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}\left(ad{d}^{\parallel a}\right)a^{\parallel d}{d}^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}\left(a{a}^{\parallel d}\right)d^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}\left(d^{\parallel a}d{d}^{\parallel a}\right){\left(a^{\parallel d}{d}^{\parallel a}\right)}^{m-1}\hfill \\ & \hfill =\hfill & d{\left(ad\right)}^{m-1}{d}^{\parallel a}{\left(a^{\parallel d}{d}^{\parallel a}\right)}^{m-1}\hfill \\ & \hfill ⋮\hfill & \\ & \hfill =\hfill & d{d}^{\parallel a}.\hfill \end{array}$$

Similarly, we have ${\left(d{\left(ad\right)}^m\right)}^{\#}d{\left(ad\right)}^m=d^{\parallel a}d$. Hence, $a{a}^{\parallel d}=d^{\parallel a}d=d{d}^{\parallel a}=a^{\parallel d}a$. □

In a ∗-ring, it is clear that $a\in R^{†}$ if and only if $a^{\ast }\in R^{\parallel a}$. Applying Theorem 5, we obtain the following well-known results (see [2] [Theorem 7.3] and [7] [Theorem 2.1]).

Corollary 3.

Let $a\in R^{†}$ and $k\in \mathbb{N}$, $m\in \mathbb{N}\cup \left\{0\right\}$. Then, the following statements are equivalent:

(i)

$a\in R^{\mathrm{EP}}$.

(ii)

$a\in R^{\#}$ and $a{a}^{\#}={\left(a{a}^{\#}\right)}^{\ast }$.

(iii)

$a\in R^{\#}$ and ${\left(a^{†}\right)}^k={\left(a^{\#}\right)}^k$.

(iv)

$a^{\ast }{\left(a{a}^{\ast }\right)}^m\in R^{\mathrm{EP}}$.

Proof. 

Let $d=a^{\ast }$ in Theorem 5. Then, the equivalences of (i), (ii) and (iii) are straightforward. So, we only need to show that (i) ⇔ (iv). Since $a\in R^{†}$, we can directly check ${\left(a^{\ast }{\left(a{a}^{\ast }\right)}^m\right)}^{†}={\left(a^{\ast }\right)}^{†}{\left(a^{†}{\left(a^{\ast }\right)}^{†}\right)}^m$. Hence,

$$\begin{array}{ccc}\hfill a\in R^{\mathrm{EP}}& \hfill ⟺\hfill & a^{\ast }{\left(a{a}^{\ast }\right)}^m\in R^{\#}and{\left(a^{\ast }{\left(a{a}^{\ast }\right)}^m\right)}^{\#}={\left(a^{\ast }\right)}^{†}{\left(a^{†}{\left(a^{\ast }\right)}^{†}\right)}^m\hfill \\ & \hfill ⟺\hfill & a^{\ast }{\left(a{a}^{\ast }\right)}^m\in R^{\#}\cap R^{†}and{\left(a^{\ast }{\left(a{a}^{\ast }\right)}^m\right)}^{\#}={\left(a^{\ast }{\left(a{a}^{\ast }\right)}^m\right)}^{†}\hfill \\ & \hfill ⟺\hfill & a^{\ast }{\left(a{a}^{\ast }\right)}^m\in R^{\mathrm{EP}}.\hfill \end{array}$$

□

The second characterization of the commutative inverse along an element is given in terms of the ring units as follows.

Theorem 6.

Let $a,d\in R$ be such that $d\in R^{\parallel a}$ and $k\in \mathbb{N}$. Then, the following statements are equivalent:

(i)

$a\in R^{\parallel d}$ and $a{a}^{\parallel d}=a^{\parallel d}a$.

(ii)

There exist $u,v\in R^{-1}$ such that $d=a^{k}u=v{a}^k$.

(iii)

There exist $u^{\prime },v^{\prime }\in R^{-1}$ such that ${\left(da\right)}^k=u^{\prime }d$ and ${\left(ad\right)}^k=d{v}^{\prime }$.

(iv)

There exist $p\in R$ and $q\in R^{-1}$ such that $pa=ap=pd=dp=0$ and $a^k=p+q$.

Proof. 

(i) ⇒ (ii) and (iii). By Theorem 5 (i) and (ii), we have $a,d\in R^{\#}$ and $a{a}^{\#}=d{d}^{\#}$. Let $u={\left(a^{\#}\right)}^{k}d+1-d{d}^{\#}$ and $v=d{\left(a^{\#}\right)}^k+1-d{d}^{\#}$. Then, we can verify that

$$u(d^{\#}{a}^k+1-d{d}^{\#})=1=(d^{\#}{a}^k+1-d{d}^{\#})u.$$

Hence, $u\in R^{-1}$, which gives that $v\in R^{-1}$ by Lemma 6. Moreover, it is obvious that $d=a^{k}u=v{a}^k$.

Similarly, let $u^{\prime }={\left(da\right)}^k{d}^{\#}+1-d{d}^{\#}$ and $v^{\prime }=d^{\#}{\left(ad\right)}^k+1-d{d}^{\#}$. Then, such $u^{\prime }$ and $v^{\prime }$ satisfy item (iii).

(ii) ⇒ (i). By the hypotheses, we obtain $a\in adaR=a^{k+1}uaR\subseteq a^{2}R$. On the other hand, $a\in R{a}^2$. So, $a\in R^{\#}$. Hence, $a=a^k{\left(a^{\#}\right)}^{k-1}=d{u}^{-1}{\left(a^{\#}\right)}^{k-1}\in dR$, together with $d=a^{k}u\in aR$, yields $aR=dR$. Similarly, $Ra=Rd$. In view of Lemma 10 (i) and (ii), we infer that item (i) holds.

(iii) ⇒ (i). Clearly, we have $a\in adaR={\left(ad\right)}^k{\left({\left(ad\right)}^{\#}\right)}^{k-1}aR=d{v}^{\prime }{\left({\left(ad\right)}^{\#}\right)}^{k-1}aR\subseteq dR$ and $d={\left(ad\right)}^k{\left(v^{\prime }\right)}^{-1}\in aR$. Hence, $aR=dR$. Dually, $Ra=Rd$, as desired.

(i) ⇒ (iv). Recall that $a,d\in R^{\#}$ and $a{a}^{\#}=d{d}^{\#}$. In view of ([25] Corollary 2.10), we obtain $q:=a^k+1-a{a}^{\#}\in R^{-1}$. Let $p=a{a}^{\#}-1$. Then, $pa=ap=pd=dp=0$ and $a^k=p+q$.

(iv) ⇒ (i). Note that $aq=qa$ and $a=aq{q}^{-1}=a(a^k-p)q^{-1}=a^{k+1}{q}^{-1}\in a^{2}R\cap R{a}^2$. Then, $a\in R^{\#}$ and $a^{\#}=a^{2k-1}{q}^{-2}=q^{-2}{a}^{2k-1}$. In addition, we deduce that

$$q^{2}d{a}^{2k}={(a^k-p)}^{2}d{a}^{2k}=a^{2k}d{a}^{2k}=a^{2k}d{(a^k-p)}^2=a^{2k}d{q}^2.$$

Hence, we have

$$da{a}^{\#}=q^{-2}\left(q^{2}d{a}^{2k}\right)q^{-2}=q^{-2}\left(a^{2k}d{q}^2\right)q^{-2}=q^{-2}{a}^{2k}d=a^{\#}ad,$$

which implies $d{d}^{\parallel a}=d^{\parallel a}d$ by Lemma 9 (i) and (ii). Since $d=q^{-1}qd=q^{-1}(p+q)d=q^{-1}{a}^{k}d$, we conclude $Rd\subseteq Rad$. Hence, by Lemma 1, we have $a\in R^{\parallel d}$. So, $a{a}^{\parallel d}=a^{\parallel d}a$. □