# Some New Results on Pseudo $n$-Strong
Drazin Inverses in Rings

Jian Cui  School of Mathematics and Statistics, Anhui Normal University
Wuhu, Anhui 241002, China<EMAIL_ADDRESS>,  Peter Danchev  Institute of
Mathematics and Informatics, Bulgarian Academy of Sciences
”Acad. G. Bonchev” str., bl. 8, 1113 Sofia, Bulgaria<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Yuedi Zeng  Key Laboratory of Financial Mathematics
(Putian University), Fujian Province University, Putian, Fujian, China
<EMAIL_ADDRESS>
###### Abstract.

In this paper, we give a further study in-depth of the pseudo $n$-strong
Drazin inverses in an associative unital ring $R$. The characterizations of
elements $a,b\in R$ for which $aa^{\tiny{\qihaoD⃝}}=bb^{\tiny{\qihaoD⃝}}$ are
provided, and some new equivalent conditions on pseudo $n$-strong Drazin
inverses are obtained. In particular, we show that an element $a\in R$ is
pseudo $n$-strong Drazin invertible if, and only if, $a$ is $p$-Drazin
invertible and $a-a^{n+1}\in\sqrt{J(R)}$ if, and only if, there exists
$e^{2}=e\in{\rm comm}^{2}(a)$ such that $ae\in\sqrt{J(R)}$ and
$1-(a+e)^{n}\in\sqrt{J(R)}$. We also consider pseudo $n$-strong Drazin
inverses with involution, and discuss the extended versions of Cline’s formula
and Jacobson’s lemma of this new class of generalized inverses. Likewise, we
define and explore the so-called pseudo $\pi$-polar rings and demonstrate
their relationships with periodic rings and strongly $\pi$-regular rings,
respectively.

###### Key words and phrases:

Drazin inverse; pseudo $n$-strong Drazin inverse; pseudo Drazin inverse;
$pns$-$*$-Drazin inverse

###### 2020 Mathematics Subject Classification:

16U50; 16W10

##  1\. Introduction and Motivation

Let $R$ be an arbitrary associative ring with unity. The symbols $U(R)$,
$J(R)$ and $R^{\rm nil}$ stand, respectively, for the set of all invertible
elements, the Jacobson radical and the set of all nilpotent elements of $R$.
We use ${\rm M}_{n}(R)$ to denote the $n\times n$ matrix ring over $R$. For
any $a\in R$, the commutant and double commutant of $a$ are defined by ${\rm
comm}(a)=\\{x\in R:ax=xa\\}$ and ${\rm comm}^{2}(a)$$=\\{x\in
R:xy=yx,{\rm~{}for~{}all}~{}y\in{\rm comm}(a)\\}$, respectively.

Recall that an element $a\in R$ is said to be _Drazin invertible_ if there
exists $x\in R$ such that

$None$ $xax=x,~{}~{}x\in{\rm comm}^{2}(a),~{}~{}a-a^{2}x\in R^{\rm nil}.$

Herein, the element $x$ is called a _Drazin inverse_ of $a$, which is unique
if it exists. It is well known that the condition “$x\in{\rm comm}^{2}(a)$” in
$(\ast)$ can be replaced with “$x\in{\rm comm}(a)$”. The notion of Drazin
inverses was initially introduced by Drazin [10] in 1958, which plays
important roles in matrix theory, Markov chains, partial differential
equations, etc. (see [2, 5, 21]). Moreover, Drazin inverses are closely
related to the well-known strongly $\pi$-regular rings in ring theory. In
fact, a ring $R$ is _strongly $\pi$-regular_ (see [23]) if, for each $a\in R$,
the relation $a^{n}\in a^{n+1}R\cap Ra^{n+1}$ holds for some positive integer
$n$ depending on $a$ (which ring is also called _polar_ in the existing
literature). It is well known that a ring $R$ is strongly $\pi$-regular
precisely when every element of $R$ is Drazin invertible.

For a ring $R,$ let us denote $\sqrt{J(R)}=\\{a\in R:a^{k}\in J(R)~{}{\rm
for~{}some}~{}k\geq 1\\}.$ It is easy to see that both $R^{\rm nil}$ and
$J(R)$ are contained in $\sqrt{J(R)}$.

In 2012, Wang and Chen [24] introduced the notion of the pseudo Drazin inverse
in rings. Indeed, an element $x\in R$ is called _pseudo Drazin invertible_
(or, for short, _$p$ -Drazin invertible_) if there exists $x\in R$ such that

$xax=x$, $x\in{\rm comm}^{2}(a)$, $a-a^{2}x\in\sqrt{J(R)}$.

Any element $x\in R$ satisfying the above conditions is also called a _pseudo
Drazin inverse_ (or, shortly, _$p$ -Drazin inverse_) of $a$; notice that the
pseudo Drazin inverse is unique once it exists. Clearly, Drazin invertible
elements are $p$-Drazin invertible. According to [24], $a\in R$ is $p$-Drazin
invertible exactly when $a$ is a pseudo-polar element (i.e., there exists
$p^{2}=p\in{\rm comm}^{2}(a)$ such that $a+p\in U(R)$ and $ap\in\sqrt{J(R)}$).
For more interesting properties of $p$-Drazin inverses, one may refer to [11,
19, 25, 26] and the reference therewith.

In 2021, Mosi$\acute{c}$ [20] introduced the concept of the pseudo $n$-strong
Drazin inverse which generalizes the concept of $p$-Drazin inverses. Let
$\mathbb{N}$ be the set of all positive integers and $n\in\mathbb{N}$. An
element $a\in R$ is called _pseudo $n$-strong Drazin invertible_ (or, briefly,
_$pns$ -Drazin invertible_) if there exists $x\in R$ such that

$xax=x$, $x\in{\rm comm}^{2}(a)$, $a^{n}-ax\in\sqrt{J(R)},$

where $x$ is a _pseudo $n$-strong Drazin inverse_ (or, for short, _$pns$
-Drazin inverse_) of $a$, which is unique when it exists, and denoted by
$a^{\tiny{\qihaoD⃝}}$. It was shown that any $pns$-Drazin invertible element
is $p$-Drazin invertible (see, for instance, [20]).

Motivated by all of the mentioned above information, we continue in the
present paper the investigation of $pns$-Drazin inverses in arbitrary
(associative) rings. To that goal, our work is organized thus: in Section 2,
the elements $a,b\in R$ for which $aa^{\tiny{\qihaoD⃝}}=bb^{\tiny{\qihaoD⃝}}$
are studied, and some equivalent characterizations are obtained. In
particular, it is established that $a\in R$ is $pns$-Drazin invertible if, and
only if, $a$ is $p$-Drazin invertible and $a-a^{n+1}\in\sqrt{J(R)}$ if, and
only if, there exists $p^{2}=p\in{\rm comm}^{2}(a)$ such that
$ap\in\sqrt{J(R)}$ and $1-(a+p)^{n}\in\sqrt{J(R)}.$ Further, Section 3 is
devoted to the exploration of $pns$-Drazin inverses with involution. Recall
that a ring $R$ is a _$*$ -ring_ (see, e.g., [3]) if there exists a map
$*:R\rightarrow R$ such that

$(x+y)^{*}=x^{*}+y^{*},~{}(xy)^{*}=y^{*}x^{*},~{}(x^{*})^{*}=x$

for all $x,y\in R$. For a $*$-ring $R$ and $n\in\mathbb{N}$, we shall say that
the element $a\in R$ is $pns$-$*$-Drazin invertible if $a$ is $pns$-Drazin
invertible and $(aa^{\tiny{\qihaoD⃝}})^{*}=aa^{\tiny{\qihaoD⃝}}$. On this
vein, some properties of $pns$-$*$-Drazin inverses are discussed too. It is
also proved that, for a $*$-ring $R$ and elements $a$, $b$, $c\in R$, if
$aba=ba^{2}=a^{2}c=aca$, then $ac$ (resp., $1-ac$) is $pns$-$*$-Drazin
invertible if, and only if, so is $ba$ (resp., $1-ba$) and, moreover, the
related formulae are provided. In our final Section 4, we deal with some
characterizing theorems pertaining to pseudo $\pi$-polar rings, that are those
rings whose elements are $pns$-Drazin invertible for various positive integers
$n$, examining their basic properties and transversal with the classical sorts
of periodic and strongly $\pi$-regular rings by showing that any pseudo
$\pi$-polar ring $R$ with nil $J(R)$ is periodic and vice versa. Some other
closely related assertions are proved as well.

##  2\. $pns$-Drazin inverses

The following easy lemma will be used freely in the context.

######  Lemma 2.1.

Let $a,b\in R$ and $ab=ba$.

$(1)$ If $a\in\sqrt{J(R)}$, then $ab\in\sqrt{J(R)}.$

$(2)$ If $a,b\in\sqrt{J(R)}$, then $a+b\in\sqrt{J(R)}$.

$(3)$ If $a\in\sqrt{J(R)},b\in U(R)$, then $a+b\in U(R)$.

$(4)$ $a\in\sqrt{J(R)}$ if, and only if, $au\in\sqrt{J(R)}$ for any $u\in
U(R)\cap{\rm comm}(a)$ if, and only if, $a^{k}\in\sqrt{J(R)}$ for any integer
$k>1$.

The next lemma appeared as [20, Theorem 3.1].

######  Lemma 2.2.

Let $a\in R$ and $n\in\mathbb{N}$. The following are equivalent$:$

$(1)$ The element $a$ is $pns$-Drazin invertible.

$(2)$ There exists $e^{2}=e\in{\rm comm}^{2}{(a)}$ such that
$a^{n}-e\in\sqrt{J(R)}.$
In this case, $e=aa^{\tiny{\qihaoD⃝}}$ and
$a^{\tiny{\qihaoD⃝}}=(1+a^{n}-e)^{-1}a^{n-1}e.$

For the sake of convenience, we use the symbol $R^{\tiny{\qihaoD⃝}}$ to denote
the set of all $pns$-Drazin invertible elements in a ring $R$. If $a\in
R^{\tiny{\qihaoD⃝}}$, we write $a^{\Pi}=aa^{\tiny{\qihaoD⃝}}$. Then,
$a^{\Pi}=(a^{\Pi})^{2}$,
$a^{\tiny{\qihaoD⃝}}=(1+a^{n}-a^{\Pi})^{-1}a^{n-1}a^{\Pi}$ and
$a^{\tiny{\qihaoD⃝}}a^{\Pi}=a^{\Pi}a^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}$.

As usual, for an element $a\in R$, we denote by $r(a)$ the right annihilator
of $a$ in $R$. Our motivating of the following result is originated from [13,
Theorem 6.1].

######  Theorem 2.3.

Let $a\in R^{\tiny{\qihaoD⃝}}$, $b\in R$ and $n\in\mathbb{N}$. The following
are equivalent$:$

$(1)$ $b\in R^{\tiny{\qihaoD⃝}}$ and $a^{\Pi}=b^{\Pi}$.

$(2)$ $a^{\Pi}\in{\rm comm}^{2}(b)$ and $b^{n}-a^{\Pi}\in\sqrt{J(R)}.$

$(3)$ $a^{\Pi}\in{\rm comm}^{2}(b)$, $b(1-a^{\Pi})\in\sqrt{J(R)}$ and
$(1-b^{n})a^{\Pi}\in\sqrt{J(R)}$.

$(4)$ $b\in R^{\tiny{\qihaoD⃝}},$ $1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b)\in U(R)$
and
$b^{\tiny{\qihaoD⃝}}=(1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b))^{-1}a^{\tiny{\qihaoD⃝}}$.

$(5)$ $b\in R^{\tiny{\qihaoD⃝}}$ and
$b^{\tiny{\qihaoD⃝}}-a^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}(a-b)b^{\tiny{\qihaoD⃝}}.$

$(6)$ $b\in R^{\tiny{\qihaoD⃝}}$, $a^{\Pi}b^{\Pi}=b^{\Pi}a^{\Pi}$ and
$1-(a^{\Pi}-b^{\Pi})^{2}\in U(R)$.

$(7)$ $b\in R^{\tiny{\qihaoD⃝}}$, $b^{\tiny{\qihaoD⃝}}R\subseteq
a^{\tiny{\qihaoD⃝}}R$ and $r(b^{\tiny{\qihaoD⃝}})\subseteq
r(a^{\tiny{\qihaoD⃝}})$.

###### Proof.

$(1)\Leftrightarrow(2)$. It follows immediately from Lemma 2.2.

$(2)\Leftrightarrow(3)$. Note that $ba^{\Pi}=a^{\Pi}b$ is implied by
$a^{\Pi}\in{\rm comm}^{2}(b)$. Therefore,

$b^{n}-a^{\Pi}\in\sqrt{J(R)}\Leftrightarrow(b^{n}-a^{\Pi})(1-a^{\Pi})\in\sqrt{J(R)},~{}(b^{n}-a^{\Pi})a^{\Pi}\in\sqrt{J(R)},$

and

$(b^{n}-a^{\Pi})(1-a^{\Pi})=b^{n}(1-a^{\Pi})\in\sqrt{J(R)}\Leftrightarrow
b(1-a^{\Pi})\in\sqrt{J(R)}.$

Furthermore,
$(b^{n}-a^{\Pi})a^{\Pi}=(b^{n}-1)a^{\Pi}\in\sqrt{J(R)}\Leftrightarrow(1-b^{n})a^{\Pi}\in\sqrt{J(R)}.$
This proves the equivalence between $(2)$ and $(3)$.

$(1)\Rightarrow(4)$. By $(1)$, $a^{\Pi}=b^{\Pi}$. Observe that
$1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b)=(1-a^{\Pi})+a^{\tiny{\qihaoD⃝}}b$.
Consequently,

$\begin{split}[(1-a^{\Pi})+a^{\tiny{\qihaoD⃝}}b][(1-a^{\Pi})+b^{\tiny{\qihaoD⃝}}a]&=(1-a^{\Pi})+a^{\tiny{\qihaoD⃝}}b(1-a^{\Pi})+(1-a^{\Pi})b^{\tiny{\qihaoD⃝}}a+a^{\tiny{\qihaoD⃝}}bb^{\tiny{\qihaoD⃝}}a\\\
&=1-a^{\Pi}+a^{\tiny{\qihaoD⃝}}(1-a^{\Pi})b+(1-b^{\Pi})b^{\tiny{\qihaoD⃝}}a+a^{\tiny{\qihaoD⃝}}b^{\Pi}a\\\
&=1-a^{\Pi}+a^{\tiny{\qihaoD⃝}}a^{\Pi}a=1.\end{split}$

We can similarly deduce that
$[(1-a^{\Pi})+b^{\tiny{\qihaoD⃝}}a][(1-a^{\Pi})+a^{\tiny{\qihaoD⃝}}b]=1$. So
$1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b)=(1-a^{\Pi})+a^{\tiny{\qihaoD⃝}}b\in U(R)$.
It follows that

$(1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b))b^{\tiny{\qihaoD⃝}}=(1-a^{\Pi}+a^{\tiny{\qihaoD⃝}}b)b^{\tiny{\qihaoD⃝}}=(1-b^{\Pi})b^{\tiny{\qihaoD⃝}}+a^{\tiny{\qihaoD⃝}}bb^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}b^{\Pi}=a^{\tiny{\qihaoD⃝}}a^{\Pi}=a^{\tiny{\qihaoD⃝}},$

which implies that
$b^{\tiny{\qihaoD⃝}}=(1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b))^{-1}a^{\tiny{\qihaoD⃝}}$.

$(4)\Rightarrow(5)$. By $(4),$ we have
$a^{\tiny{\qihaoD⃝}}=(1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b))b^{\tiny{\qihaoD⃝}}$.
Thus,

$b^{\tiny{\qihaoD⃝}}-a^{\tiny{\qihaoD⃝}}=b^{\tiny{\qihaoD⃝}}-(1-a^{\Pi}(1-a^{\tiny{\qihaoD⃝}}b))b^{\tiny{\qihaoD⃝}}=(a^{\Pi}-a^{\tiny{\qihaoD⃝}}b)b^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}(a-b)b^{\tiny{\qihaoD⃝}}.$

$(5)\Rightarrow(1)$. Since
$b^{\tiny{\qihaoD⃝}}-a^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}(a-b)b^{\tiny{\qihaoD⃝}},$
one has $b^{\tiny{\qihaoD⃝}}=a^{\tiny{\qihaoD⃝}}[1+(a-b)b^{\tiny{\qihaoD⃝}}]$
and $a^{\tiny{\qihaoD⃝}}=[1-a^{\tiny{\qihaoD⃝}}(a-b)]b^{\tiny{\qihaoD⃝}}$.
Then,
$(1-a^{\Pi})b^{\tiny{\qihaoD⃝}}=(1-a^{\Pi})a^{\tiny{\qihaoD⃝}}[1+(a-b)b^{\tiny{\qihaoD⃝}}]=0$
and
$a^{\tiny{\qihaoD⃝}}(1-b^{\Pi})=[1-a^{\tiny{\qihaoD⃝}}(a-b)]b^{\tiny{\qihaoD⃝}}(1-b^{\Pi})=0$,
it follows that $0=(1-a^{\Pi})b^{\tiny{\qihaoD⃝}}b=(1-a^{\Pi})b^{\Pi}$ and
$0=aa^{\tiny{\qihaoD⃝}}(1-b^{\Pi})=a^{\Pi}(1-b^{\Pi})$. Therefore,
$a^{\Pi}=a^{\Pi}b^{\Pi}=b^{\Pi}$.

$(1)\Rightarrow(6)$. It is clear.

$(6)\Rightarrow(1)$. From $a^{\Pi}b^{\Pi}=b^{\Pi}a^{\Pi}$, we have
$(a^{\Pi}-b^{\Pi})=(a^{\Pi}-b^{\Pi})^{3}$. So $1-(a^{\Pi}-b^{\Pi})^{2}$ is an
idempotent. By hypothesis, $1-(a^{\Pi}-b^{\Pi})^{2}\in U(R).$ Thus,
$1-(a^{\Pi}-b^{\Pi})^{2}=1,$ which implies $(a^{\Pi}-b^{\Pi})^{2}=0$, and
hence $a^{\Pi}-b^{\Pi}=(a^{\Pi}-b^{\Pi})^{3}=0.$ So, the result follows.

$(1)\Rightarrow(7)$. Since $a^{\Pi}=b^{\Pi},$ we obtain
$b^{\tiny{\qihaoD⃝}}R=(b^{\Pi}b^{\tiny{\qihaoD⃝}})R=(a^{\Pi}b^{\tiny{\qihaoD⃝}})R=(a^{\tiny{\qihaoD⃝}}ab^{\tiny{\qihaoD⃝}})R\subseteq
a^{\tiny{\qihaoD⃝}}R$. Similarly,
$Ra^{\tiny{\qihaoD⃝}}=R(a^{\tiny{\qihaoD⃝}}a^{\Pi})=R(a^{\tiny{\qihaoD⃝}}b^{\Pi})=R(a^{\tiny{\qihaoD⃝}}b^{\Pi}b^{\tiny{\qihaoD⃝}})\subseteq
Rb^{\tiny{\qihaoD⃝}}$, which forces that
$r(b^{\tiny{\qihaoD⃝}})=r(Rb^{\tiny{\qihaoD⃝}})\subseteq
r(Ra^{\tiny{\qihaoD⃝}})=r(a^{\tiny{\qihaoD⃝}})$.

$(7)\Rightarrow(1)$. One observes that
$b^{\tiny{\qihaoD⃝}}=b^{\tiny{\qihaoD⃝}}bb^{\tiny{\qihaoD⃝}}$ and
$b^{\tiny{\qihaoD⃝}}b=b^{\Pi}$. Then,
$b^{\tiny{\qihaoD⃝}}R=b^{\tiny{\qihaoD⃝}}bR=b^{\Pi}R$ and
$Rb^{\tiny{\qihaoD⃝}}=Rbb^{\tiny{\qihaoD⃝}}=Rb^{\Pi}.$ We can prove similarly
that $a^{\tiny{\qihaoD⃝}}R=a^{\tiny{\qihaoD⃝}}aR=a^{\Pi}R$ and
$Ra^{\tiny{\qihaoD⃝}}=Raa^{\tiny{\qihaoD⃝}}=Ra^{\Pi}.$ Since
$b^{\tiny{\qihaoD⃝}}R\subseteq a^{\tiny{\qihaoD⃝}}R$, we get
$b^{\Pi}R\subseteq a^{\Pi}R$, which gives $b^{\Pi}=a^{\Pi}b^{\Pi}$. Note that
$r(b^{\tiny{\qihaoD⃝}})=r(Rb^{\tiny{\qihaoD⃝}})=r(Rb^{\Pi})=(1-b^{\Pi})R$ and
$r(a^{\tiny{\qihaoD⃝}})=r(Ra^{\tiny{\qihaoD⃝}})=r(Ra^{\Pi})=(1-a^{\Pi})R$.
Since $r(b^{\tiny{\qihaoD⃝}})\subseteq r(a^{\tiny{\qihaoD⃝}})$, we have
$(1-b^{\Pi})R\subseteq(1-a^{\Pi})R,$ which yields
$1-b^{\Pi}=(1-a^{\Pi})(1-b^{\Pi}).$ Thus, $a^{\Pi}=a^{\Pi}b^{\Pi}$ and,
therefore, $a^{\Pi}=b^{\Pi}$. ∎

As a consequence, we derive the following.

######  Corollary 2.4.

Let $a\in R$, $e^{2}=e\in R$ and $n\in\mathbb{N}$. The following are
equivalent$:$

$(1)$ $a$ is $pns$-Drazin invertible with $a^{\Pi}=e$.

$(2)$ $e^{2}=e\in{\rm comm}^{2}{(a)}$, $a(1-e)\in\sqrt{J(R)}$ and
$(1-a^{n})e\in\sqrt{J(R)}.$

###### Proof.

Given any $f^{2}=f\in R,$ we can check that $f\in R^{\tiny{\qihaoD⃝}}$ and
$f^{\tiny{\qihaoD⃝}}=f$, whence $f^{\Pi}=ff^{\tiny{\qihaoD⃝}}=f$.

$(1)\Rightarrow(2)$. Since $e^{2}=e\in R$, one sees that $e\in
R^{\tiny{\qihaoD⃝}}$ with $e^{\Pi}=e$. So, we have $a^{\Pi}=e^{\Pi}$. In view
of Theorem 2.3, $e^{2}=e\in{\rm comm}^{2}{(a)}$, $a(1-e)\in\sqrt{J(R)}$ and
$(1-a^{n})e\in\sqrt{J(R)}$.

$(2)\Rightarrow(1)$. As $e=e^{\Pi}$, it follows at once from Theorem 2.3 that
$a\in R^{\tiny{\qihaoD⃝}}$ and $a^{\Pi}=e^{\Pi}=e$. ∎

By virtue of Corollary 2.4, we extract the following assertion.

######  Corollary 2.5.

Let $a\in R$ and $n\in\mathbb{N}$. Then,

$(1)$ $a$ is $pns$-Drazin invertible with $a^{\Pi}=1$ if, and only if,
$a^{n}-1\in\sqrt{J(R)}.$

$(2)$ $a$ is $pns$-Drazin invertible with $a^{\Pi}=0$ if, and only if,
$a\in\sqrt{J(R)}.$

Recall again that an element $a\in R$ is _$p$ -Drazin invertible_ if there
exists $y\in R$ such that

$yax=y,~{}~{}y\in{\rm comm}^{2}(a)~{}~{}{\rm and}~{}a-a^{2}y\in\sqrt{J(R)},$

where $y$ is called a $p$-Drazin inverse of $a$.

The following result establishes the relationship between a $pns$-Drazin
invertible element and a $p$-Drazin invertible element in a ring.

######  Theorem 2.6.

Let $a\in R$ and $n\in\mathbb{N}$. The following are equivalent$:$

$(1)$ The element $a$ is $pns$-Drazin invertible.

$(2)$ There exists a unique $e^{2}=e\in{\rm comm}{(a)}$ such that
$a^{n}-e\in\sqrt{J(R)}.$

$(3)$ There exists a unique $y\in R$ such that $ya=ay$, $y=yay$ and
$a^{n}-ay\in\sqrt{J(R)}.$

$(4)$ $a$ is $p$-Drazin invertible and $a-a^{n+1}\in\sqrt{J(R)}$.

###### Proof.

$(1)\Rightarrow(4)$. Clearly, $pns$-Drazin invertible elements are $p$-Drazin
invertible. In view of Lemma 2.2, there exists $e^{2}=e\in{\rm comm}^{2}{(a)}$
such that $a^{n}-e\in\sqrt{J(R)}$. As $ae=ea,$ we have

$a^{n+1}-ae=a(a^{n}-e)\in\sqrt{J(R)},$

and

$[a(1-e)]^{n}=(a^{n}-e)(1-e)\in\sqrt{J(R)},$

which implies $a(1-e)\in\sqrt{J(R)}.$ It now follows that

$a-a^{n+1}=a(1-e)-(a^{n+1}-ae)\in\sqrt{J(R)},$

as wanted.

$(4)\Rightarrow(2)$. Since $a$ is $p$-Drazin invertible, with [24, Theorem
3.2] at hand there exists $p^{2}=p\in{\rm comm}^{2}(a)$ such that $a+p\in
U(R)$ and $ap\in\sqrt{J(R)}$. Write $u=a+p$ and $e=1-p$. Then, $ep=pe=0$ and
$ae=ue=eu=ea$. Since $a-a^{n+1}\in\sqrt{J(R)}$, it must be that

$(a-a^{n+1})e=(u-u^{n+1})e\in\sqrt{J(R)},$

which implies that $(1-u^{n})e\in\sqrt{J(R)},$ and so

$(a^{n}-e)e=u^{n}e-e=(u^{n}-1)e\in\sqrt{J(R)}.$

As $ap\in\sqrt{J(R)}$, we get $(a^{n}-e)(1-e)=(ap)^{n}\in\sqrt{J(R)}.$
Therefore,

$a^{n}-e=(a^{n}-e)e+(a^{n}-e)(1-e)\in\sqrt{J(R)}.$

To show that the idempotent $e$ above is unique, we assume that there exists
$f^{2}=f\in{\rm comm}(a)$ satisfying $a^{n}-f\in\sqrt{J(R)}.$ Set $a^{n}-f=b.$
Then, $a^{n}f=(f+b)f=(2f-1+b)f$, where $2f-1+b\in R^{-1}$ as $(2f-1)^{2}=1$
and $fb=bf$. Thus, $f=(2f-1+b)^{-1}a^{n}f$. From $e\in{\rm comm}^{2}(a)$, one
has $(1-e)f$ is an idempotent. However,

$(1-e)f=(1-e)[(2f-1+b)^{-1}a^{n}f]=(a^{n}-e)(1-e)(2f-1+b)^{-1}f\in\sqrt{J(R)}$

since $a^{n}-e\in\sqrt{J(R)}$ and $e,f,a,b$ commute with one another. This
proves that $(1-e)f=0$. Similarly, we can show that $(1-f)e=0$. Hence,
$e=fe=ef=f,$ as desired.

$(2)\Rightarrow(3)$. Assume that $(2)$ holds. Put
$y=(1+a^{n}-e)^{-1}a^{n-1}e$. Then, one inspects that

$ya=(1+a^{n}-e)^{-1}a^{n}e=(1+a^{n}-e)^{-1}(a^{n}+1-e)e=e=ay,$

$yay=ye=y$ and $a^{n}-ay=a^{n}-e\in\sqrt{J(R)}$. Suppose now that there exists
another $z\in R$ such that $za=az$, $z=zaz$ and $a^{n}-az\in\sqrt{J(R)}$. Let
$f=az.$ Then, $f\in{\rm comm}(a)$ and $a^{n}-f=a^{n}-az\in\sqrt{J(R)}$. By
hypothesis, $e=f.$ So, $az=za=ya=ay,$ which assures $y=yay=zay=zaz=z$, as
expected.

$(3)\Rightarrow(2)$. This implication follows in a similar manner to that of
$(2)\Rightarrow(3)$.

$(2)\Rightarrow(1)$. Using Lemma 2.2, it suffices to show that $e\in{\rm
comm}^{2}(a)$. For any $x\in{\rm comm}(a),$ set $p=e+ex(1-e)$. Evidently,
$p^{2}=p$ and $p\in{\rm comm}(a)$. Since $a^{n}-e=:c\in\sqrt{J(R)},$ we obtain
$a^{n}=(1-e)+(2e-1+c)$ and $(a^{n}-e)^{k}\in J(R)$ for some integer $k$. It,
therefore, follows that $a^{n}e=(2e-1+c)e$ and
$a^{nk}(1-e)=(a^{n}-e)^{k}(1-e)\in J(R).$ Write $v=2e-1+c$. Then, $v\in U(R)$
and $e=v^{-1}a^{n}e.$ Since $ax=xa$ and $a^{nk}(1-e)\in J(R),$ we conclude
that

$ex(1-e)=e^{k}x(1-e)=v^{-k}a^{nk}ex(1-e)=v^{-k}exa^{nk}(1-e)\in J(R),$

which insures that

$a^{n}-p=(a^{n}-e)-ex(1-e)\in\sqrt{J(R)}.$

By assumption, $e=p$. Thus, we obtain $ex=exe.$ Similarly, we can obtain
$xe=exe$. Therefore, $ex=xe$, and so $e\in{\rm comm}^{2}(a)$, as required. ∎

By the usage of [20, Theorem 3.1], [24, Theorem 3.2] and the proof of Theorem
2.6, we provide a transparent connection between a $p$-Drazin inverse and a
$pns$-Drazin inverse in a ring as follows.

######  Corollary 2.7.

Let $a,y\in R$, $e=ay$ and $n\in\mathbb{N}$. The following are equivalent$:$

$(1)$ $y\in{\rm comm}^{2}(a)$, $yay=y$ and $a^{n}-ay\in\sqrt{J(R)}$ $($i.e.,
$a\in R^{\tiny{\qihaoD⃝}}$ with $a^{\tiny{\qihaoD⃝}}=y)$.

$(2)$ $e^{2}=e\in{\rm comm}^{2}(a)$ and $a^{n}-e\in\sqrt{J(R)}$.

$(3)$ $y\in{\rm comm}^{2}(a)$, $yay=y$, $a-a^{2}y\in\sqrt{J(R)}$ and
$a-a^{n+1}\in\sqrt{J(R)}$.

$(4)$ $(1-e)^{2}=1-e\in{\rm comm}^{2}(a)$, $a+(1-e)\in U(R)$,
$a(1-e)\in\sqrt{J(R)}$ and $a-a^{n+1}\in\sqrt{J(R)}$.

######  Example 2.8.

Let $F$ be a field and $R={\rm M}_{k}(F)$. Then, $\sqrt{J(R)}=R^{\rm nil}$.
One infers from [24, Theorem 2.1] that all matrices in $R$ are $p$-Drazin
invertible. But, a simple check shows that any matrix $A\in R$ satisfying the
relation $A-A^{n+1}\notin R^{\rm nil}$ is not $pns$-Drazin invertible, as
needed.

######  Corollary 2.9.

Let $a\in R$ and $n\in\mathbb{N}$. The following are equivalent$:$

$(1)$ The element $a$ is $pns$-Drazin invertible.

$(2)$ There exists $p^{2}=p\in{\rm comm}^{2}(a)$ such that $ap\in\sqrt{J(R)}$
and $1-(a+p)^{n}\in\sqrt{J(R)}.$

###### Proof.

$(1)\Rightarrow(2)$. Let $e=aa^{\tiny{\qihaoD⃝}}$ and $p=1-e$. In virtue of
Corollary 2.7, one has that $p^{2}=p\in{\rm comm}^{2}(a)$, $a+p\in U(R)$,
$ap\in\sqrt{J(R)}$ and $a-a^{n+1}\in\sqrt{J(R)}$. Write $u=a+p$. By the
classical Binomial Theorem, we may write

$u-u^{n+1}=(a+p)-(a+p)^{n+1}=(a-a^{n+1})-ap\cdot\sum_{i=1}^{n}C_{n+1}^{i}(ap)^{n-i}\in\sqrt{J(R)}.$

Since $u\in U(R)$, it follows that $1-u^{n}\in\sqrt{J(R)}$, as required.

$(2)\Rightarrow(1)$. From $1-(a+p)^{n}\in\sqrt{J(R)},$ one has
$(a+p)-(a+p)^{n+1}\in\sqrt{J(R)}$, and $a+p\in U(R)$ as $(a+p)^{n}\in
1+\sqrt{J(R)}\subseteq U(R).$ Utilizing Corollary 2.7, we only need to show
$a-a^{n+1}\in\sqrt{J(R)}$. Similarly to the above proof, we can get

$a-a^{n+1}=[(a+p)-(a+p)^{n+1}]+ap\cdot\sum_{i=1}^{n}C_{n+1}^{i}(ap)^{n-i}\in\sqrt{J(R)}$

since $ap\in\sqrt{J(R)}.$ Thus, $(1)$ follows, as promised. ∎

##  3\. $pns$-Drazin inverses with involution

This section focuses on the study of a class of $pns$-Drazin inverses in a
$*$-ring. Recall once again that a ring $R$ is a _$*$ -ring_ (or a _ring with
involution_) if there is a map $*:R\rightarrow R$ such that for all $x,~{}y\in
R,$ $(x+y)^{*}=x^{*}+y^{*},~{}(xy)^{*}=y^{*}x^{*}$ and $(x^{*})^{*}=x$.

######  Definition 3.1.

Let $R$ be a $*$-ring and $n\in\mathbb{N}$. An element $a\in R$ is called
$pns$-$*$-Drazin invertible if there exists $x\in R$ such that

$x\in{\rm comm}^{2}(a)$, $xax=x$, $(ax)^{\ast}=ax$, $a^{n}-ax\in\sqrt{J(R)}.$

In this case, $x$ is called a $pns$-$*$-Drazin inverse of $a$.

For a $*$-ring $R$ and $n\in\mathbb{N}$. The $pns$-$*$-Drazin inverse of $a\in
R$ will be denoted by $a^{\diamond}$; and the set of all $pns$-$*$-Drazin
invertible elements of $R$ is denoted by $R^{\diamond}$. Apparently,
$R^{\diamond}\subseteq R^{\tiny{\qihaoD⃝}}$.

######  Remark 3.2.

It is clear that $a\in R^{\diamond}$ if, and only if, $a\in
R^{\tiny{\qihaoD⃝}}$ and $(aa^{\tiny{\qihaoD⃝}})^{*}=aa^{\tiny{\qihaoD⃝}}$.
So, by the uniqueness of the $pns$-Drazin inverse, one gets that the
$pns$-$*$-Drazin inverse is unique whenever it exists.

Recall that an element $p$ of a $*$-ring $R$ is a _projection_ (cf. [3]) if
$p$ is a self-adjoint idempotent (i.e., $p^{2}=p$ and $p=p^{*}$).

######  Proposition 3.3.

Let $R$ be a $*$-ring, $a\in R$ and $n\in\mathbb{N}$. The following are
equivalent$:$

$(1)$ The element $a$ is $pns$-$*$-Drazin invertible.

$(2)$ There exists a projection $p\in{\rm comm}^{2}{(a)}$ such that
$a^{n}-p\in\sqrt{J(R)}.$

###### Proof.

$(1)\Rightarrow(2)$. Clearly, $a$ is $pns$-Drazin invertible. Write
$p=aa^{\tiny{\qihaoD⃝}}$. Employing Lemma 2.2, we have $p^{2}=p\in{\rm
comm}^{2}{(a)}$ and $a^{n}-p\in\sqrt{J(R)}$. As
$(aa^{\tiny{\qihaoD⃝}})^{*}=aa^{\tiny{\qihaoD⃝}}$, we infer that $p$ is a
projection.

$(2)\Rightarrow(1)$. Let $x=(1+a^{n}-p)^{-1}a^{n-1}p.$ Applying Lemma 2.2, we
conclude that $x$ is the $pns$-Drazin inverse of $a$. Notice that

$ax=a^{n}(1+a^{n}-p)^{-1}p=(a^{n}+1-p)(1+a^{n}-p)^{-1}p=p.$

But, since $p$ is a projection, it follows that $(ax)^{*}=ax$. So, $x$ is the
$pns$-$*$-Drazin inverse of $a$, as pursued. ∎

######  Example 3.4.

Let $R$ be a $*$-ring and $n\in\mathbb{N}$.

$(1)$ Elements of $\sqrt{J(R)}$ are $pns$-$*$-Drazin invertible. In
particular, nilpotents and elements of $J(R)$ are $pns$-$*$-Drazin invertible.

$(2)$ Idempotents of $R$ are $pns$-Drazin invertible, and an idempotent $e$ is
$pns$-$*$-Drazin invertible if, and only if, $e$ is a projection.

###### Proof.

$(1)$ As $0$ is a projection, the result follows directly from Proposition
3.3.

$(2)$ Idempotents are obviously $pns$-Drazin invertible. Further, assume that
$e$ is $pns$-$*$-Drazin invertible. Then, owing to Proposition 3.3, there
exists a projection $p\in{\rm comm}^{2}(e)$ such that
$e^{n}-p=e-p\in\sqrt{J(R)}$. However, $ep=pe$ ensures $(e-p)^{3}=e-p$. Thus,
$e=p$. The converse is easy to be obtained. ∎

On the other vein, Cline [7] showed that, for any $a,b$ of a ring $R$, the
product $ab$ is Drazin invertible if, and only if, $ba$ is Drazin invertible
and $(ba)^{D}=b((ab)^{D})^{2}a$, where $(ba)^{D}$ (resp., $(ab)^{D}$) is the
Drazin inverse of $ba$ (resp., $ab$). This equation is known as the Cline’s
formula. That formula was studied for various aspects of generalized inverses
in e.g., [15, 18, 20, 24]; see also the bibliography therewith.

######  Lemma 3.5.

[20, Corollary3.2] Let $R$ be a ring and $a$, $b$, $c\in R$. If $aba=aca$,
then $ac\in R^{\tiny{\qihaoD⃝}}$ if, and only if, $ba\in R^{\tiny{\qihaoD⃝}}$.
In this case, $(ac)^{\tiny{\qihaoD⃝}}=a[(ba)^{\tiny{\qihaoD⃝}}]^{2}c$ and
$(ba)^{\tiny{\qihaoD⃝}}=b[(ac)^{\tiny{\qihaoD⃝}}]^{2}a$.

The next construction reveals that the Cline’s formula does not hold for
$pns$-$*$-Drazin inverses.

######  Example 3.6.

Let $R={\rm M}_{2}(S)$, where $S$ is a commutative ring. An involution $*$ of
$R$ is defined by the transpose of matrices. It is clear that $R$ is a
$*$-ring. Let

$A=\left(\begin{array}[]{cc}0&0\\\ 1&1\end{array}\right)$,
$B=\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right)\in R$.

Then, $AB=\left(\begin{array}[]{cc}0&0\\\ 0&1\end{array}\right)$ and
$BA=\left(\begin{array}[]{cc}1&1\\\ 0&0\end{array}\right)$ are both
idempotents. Since $AB$ is a projection of $R$, by Example 3.4(2) we deduce
that $AB$ is $pns$-$*$-Drazin invertible. However, $BA$ is not
$pns$-$*$-Drazin invertible, because it is not a projection.

We are now prepared to establish the following statement.

######  Theorem 3.7.

Let $R$ be a $*$-ring and $a$, $b$, $c\in R$. If $aba=ba^{2}=a^{2}c=aca$, then
$ac\in R^{\diamond}$ if, and only if, $ba\in R^{\diamond}$. In this case,
$(ac)^{\diamond}=a[(ba)^{\diamond}]^{2}c$ and
$(ba)^{\diamond}=b[(ac)^{\diamond}]^{2}a$.

###### Proof.

$(\Rightarrow)$. Let $y=b[(ac)^{\diamond}]^{2}a$. Exploiting Lemma 3.5, one
derives that $y(ba)y=y$, $y\in{\rm comm}^{2}(ba)$, and
$(ba)^{n}-(ba)y\in\sqrt{J(R)}$. Since $a(ba)=(ba)a=a(ac)=(ac)a$, we obtain

$\begin{split}bay=bab[(ac)^{\diamond}]^{2}a&=bab\cdot
ac[(ac)^{\diamond}]^{3}a=b(aca)c[(ac)^{\diamond}]^{3}a\\\
&=b(ac)^{\diamond}a=ba(ac)^{\diamond}=ba\cdot ac[(ac)^{\diamond}]^{2}\\\
&=(aca)c[(ac)^{\diamond}]^{2}=ac(ac)^{\diamond}.\end{split}$

Therefore, $(bay)^{\ast}=bay$, and thus, $y=ba^{\diamond}$.

$(\Leftarrow)$. Let $z=a[(ba)^{\diamond}]^{2}c$. According to Lemma 3.5, we
have $z(ac)z=z$, $z\in{\rm comm}^{2}(ac)$, and $(ac)^{n}-(ac)z\in\sqrt{J(R)}$.
To show that $ac\in R^{\diamond}$, it suffices to prove $(acz)^{\ast}=acz$. As
$a(ba)=(ba)a=a(ac)=(ac)a$, it follows that

$acz=aba[(ba)^{\diamond}]^{2}c=a(ba)^{\diamond}c=(ba)^{\diamond}ac=[(ba)^{\diamond}]^{2}ba\cdot
ac=[(ba)^{\diamond}]^{2}b(aba)=(ba)^{\diamond}ba$.

So, $(acz)^{\ast}=acz$, as required. ∎

Putting “$b=c$” in Theorem 3.7, we yield the following result immediately.

######  Corollary 3.8.

Let $R$ be a $*$-ring and $a$, $b\in R$. If $aba=ba^{2}=a^{2}b$, then $ab\in
R^{\diamond}$ if, and only if, $ba\in R^{\diamond}$.

Let $R$ be a ring and $a,b,c\in R$. The standard Jacobson’s lemma states that
if $1-ab\in U(R)$, then $1-ba\in U(R)$ and $(1-ba)^{-1}=1+b(1-ab)^{-1}a$.
However, Corach et al. [8] extended the classical Jacobson’s lemma to the case
“$aba=aca$”, and proved that if $1-ac\in U(R)$, then $1-ba\in U(R)$.

We now have the following statement for $pns$-Drazin inverses which somewhat
improves the corresponding results from [14] and [22].

######  Lemma 3.9.

Let $R$ be a ring, $a$, $b$, $c\in R$ and $n\in\mathbb{N}$. If $aba=aca$, then
$\alpha=1-ba$ is $pns$-Drazin inverses if, and only if, so is $\beta=1-ac$. In
this case,

$\alpha^{\tiny{\qihaoD⃝}}=1-b\beta^{\pi}v(ac)^{2}a+b\beta\beta^{\tiny{\qihaoD⃝}}a+b\beta^{\tiny{\qihaoD⃝}}aba$,

and

$\beta^{\tiny{\qihaoD⃝}}=1-a\alpha^{\pi}u(ba)^{2}c+a\alpha\alpha^{\tiny{\qihaoD⃝}}c+a\alpha^{\tiny{\qihaoD⃝}}cac$,

where $\alpha^{\pi}=1-\alpha\alpha^{\tiny{\qihaoD⃝}}$,
$\beta^{\pi}=1-\beta\beta^{\tiny{\qihaoD⃝}}$,
$v=(1-\beta\beta^{\pi}\sum_{i=0}^{2}(ac)^{i})^{-1}$ and
$u=(1-\alpha\alpha^{\pi}\sum_{i=0}^{2}(ba)^{i})^{-1}$.

###### Proof.

It is enough to prove only one direction, as the other direction can be showed
in a similar manner. So, we suppose that $\alpha=1-ba\in R^{\tiny{\qihaoD⃝}}$.
Let $\alpha^{\pi}=1-\alpha\alpha^{\tiny{\qihaoD⃝}}$. In accordance with
Corollary 2.7, $\alpha\alpha^{\pi}\in\sqrt{J(R)}$. So,
$t=1-\alpha\alpha^{\pi}\sum_{i=0}^{2}(ba)^{i}\in U(R)$. Let $u=t^{-1}$ and
$x=1-a\alpha^{\pi}u(ba)^{2}c+a\alpha\alpha^{\tiny{\qihaoD⃝}}c+a\alpha^{\tiny{\qihaoD⃝}}cac$.
Note that $(ba)u=u(ba)$. To show that $\beta=1-ac\in R^{\tiny{\qihaoD⃝}}$, we
next prove the following three things: (1) $x\in{\rm comm}^{2}(\beta)$; (2)
$x\beta x=x$; (3) $\beta^{n}-x\beta\in\sqrt{J(R)}$.

(1) Let $y\in R$ with $y\beta=\beta y$. Then, $ac$ commutes with $y$. So, we
obtain

(1.1) $(bacya)\alpha=bacya-bacacya=\alpha(bacya),$

which implies that $bacya$ commutes with $\alpha^{\tiny{\qihaoD⃝}}$,
$\alpha^{\pi}$, $ba$ and $u$. Notice that
$\alpha^{\pi}u^{-1}=\alpha^{\pi}-\alpha^{\pi}(1-(ba)^{3})=\alpha^{\pi}(ba)^{3}$,
and hence we have

(1.2) $\alpha^{\pi}=\alpha^{\pi}(ba)^{3}u.$

By (1.1), we obtain

(1.3)
$ya\alpha^{\pi}u(ba)^{2}c=ya(ba)^{2}\alpha^{\pi}uc=a(bacya)\alpha^{\pi}uc=a\alpha^{\pi}u(bacya)c=a\alpha^{\pi}u(ba)^{2}cy.$

Multiplying (1.3) by $ac$ on the right, we get
$ya\alpha^{\pi}c=a\alpha^{\pi}cy$ with the aid of (1.2), which yields

(1.4) $ya\alpha\alpha^{\tiny{\qihaoD⃝}}c=a\alpha\alpha^{\tiny{\qihaoD⃝}}cy.$

It now follows that

(1.5)
$yaba\alpha\alpha^{\tiny{\qihaoD⃝}}cac=aca\alpha\alpha^{\tiny{\qihaoD⃝}}cyac=aba\alpha\alpha^{\tiny{\qihaoD⃝}}cacy$

by multiplying $ac$ on both sides of (1.4). Besides, in view of (1.1),

$y(a(ba)^{2}\alpha^{\tiny{\qihaoD⃝}}cac)=ya(ba)^{3}\alpha^{\tiny{\qihaoD⃝}}c=a(bacya)ba\alpha^{\tiny{\qihaoD⃝}}c=aba\alpha^{\tiny{\qihaoD⃝}}(bacya)c=(a(ba)^{2}\alpha^{\tiny{\qihaoD⃝}}cac)y,$

which gives

(1.6)
$y(aba(1-\alpha)\alpha^{\tiny{\qihaoD⃝}}cac)=(aba(1-\alpha)\alpha^{\tiny{\qihaoD⃝}}cac)y.$

Combining (1.5) with (1.6), we obtain

(1.7) $yaba\alpha^{\tiny{\qihaoD⃝}}cac=aba\alpha^{\tiny{\qihaoD⃝}}cacy.$

Now, multiplying (1.4) by $ac$ on the right, we have
$ya\alpha\alpha^{\tiny{\qihaoD⃝}}cac=a\alpha\alpha^{\tiny{\qihaoD⃝}}cacy$, and
then
$ya(1-ba)\alpha^{\tiny{\qihaoD⃝}}cac=a(1-ba)\alpha^{\tiny{\qihaoD⃝}}cacy$. By
virtue of (1.7), we have

(1.8) $ya\alpha^{\tiny{\qihaoD⃝}}cac=a\alpha^{\tiny{\qihaoD⃝}}cacy.$

According to (1.3) (1.4) and (1.8), we get $yx=xy$, which implies that
$x\in{\rm comm}^{2}(\beta)$.

(2) Taking $y=ac$ in (1.8), one has that
$(a\alpha^{\tiny{\qihaoD⃝}}cac)ac=ac(a\alpha^{\tiny{\qihaoD⃝}}cac)$. Then, we
can calculate that

$\displaystyle x\beta=$ $\displaystyle
1-ac-a\alpha^{\pi}u(ba)^{2}c+a\alpha^{\pi}c+a\alpha\alpha^{\tiny{\qihaoD⃝}}c-a\alpha\alpha^{\tiny{\qihaoD⃝}}cac+a\alpha^{\tiny{\qihaoD⃝}}cac-a\alpha^{\tiny{\qihaoD⃝}}cacac$
$\displaystyle=$ $\displaystyle
1-a\alpha^{\pi}u(ba)^{2}c+(-ac+a\alpha^{\pi}c+a(1-a^{\pi})c)+(-a\alpha\alpha^{\tiny{\qihaoD⃝}}cac+a\alpha^{\tiny{\qihaoD⃝}}cac)-a\alpha^{\tiny{\qihaoD⃝}}cacac$
$\displaystyle=$ $\displaystyle
1-a\alpha^{\pi}u(ba)^{2}c+aba\alpha^{\tiny{\qihaoD⃝}}cac-
aca\alpha^{\tiny{\qihaoD⃝}}cac$ $\displaystyle=$ $\displaystyle
1-a\alpha^{\pi}u(ba)^{2}c.$

We, however, see that $(a\alpha^{\pi}u(ba)^{2}c)^{2}=a\alpha^{\pi}u(ba)^{2}c$
and $\alpha^{\pi}\alpha^{\tiny{\qihaoD⃝}}=0.$ So,

$x\beta
x=(1-a\alpha^{\pi}u(ba)^{2}c)(1-a\alpha^{\pi}u(ba)^{2}c+a\alpha\alpha^{\tiny{\qihaoD⃝}}c+a\alpha^{\tiny{\qihaoD⃝}}cac)=x.$

(3) Observe that

$\alpha^{n}-\alpha\alpha^{\tiny{\qihaoD⃝}}=(1-ba)^{n}-1+1-\alpha\alpha^{\tiny{\qihaoD⃝}}=\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ba)^{i}+\alpha^{\pi}$.

Since $x\beta=1-a\alpha^{\pi}u(ba)^{2}c$, we obtain

$\beta^{n}-x\beta=(1-ac)^{n}-1+a\alpha^{\pi}u(ba)^{2}c=a(\alpha^{\pi}u(ba)^{2}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c).$

Hence, one computes that

$\displaystyle(\beta^{n}-x\beta)^{2}$ $\displaystyle=$ $\displaystyle
a[\alpha^{\pi}+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i}][\alpha^{\pi}u(ba)^{2}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c]$
$\displaystyle=$ $\displaystyle
a[\alpha^{\pi}u(ba)^{2}c+\alpha^{\pi}\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ba)^{i}\alpha^{\pi}u(ba)^{2}c$
$\displaystyle+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ba)^{i}\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c]$
$\displaystyle=$ $\displaystyle
a[\alpha^{\pi}+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ba)^{i}][\alpha^{\pi}u(ba)^{2}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c]$
$\displaystyle=$ $\displaystyle
a(\alpha^{n}-\alpha\alpha^{\tiny{\qihaoD⃝}})[\alpha^{\pi}u(ba)^{2}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c].$

Repeating the same procedure as that alluded to above, we have

$(\beta^{n}-x\beta)^{m+1}=a(\alpha^{n}-\alpha\alpha^{\tiny{\qihaoD⃝}})^{m}(\alpha^{\pi}u(ba)^{2}c+\sum_{i=1}^{n}(-1)^{i}C^{i}_{n}(ca)^{i-1}c)$

for any integer $m>1.$ As
$\alpha^{n}-\alpha\alpha^{\tiny{\qihaoD⃝}}\in\sqrt{J(R)},$ one must get
$(\beta^{n}-x\beta)^{k}\in J(R)$ for some integer $k$, which implies that
$\beta^{n}-x\beta\in\sqrt{J(R)}$, as required. ∎

The next comments are worthy of mentioning.

######  Remark 3.10.

$(1)$ Jacobson’s lemma does not hold for $pns$-$*$-Drazin inverses. To show
that, let $R$ be the $*$-ring as given in Example 3.6, and let
$A=\left(\begin{array}[]{cc}0&0\\\ 1&1\end{array}\right)$,
$B=\left(\begin{array}[]{cc}0&1\\\ 0&0\end{array}\right)\in R$. Then,
$1-AB=\left(\begin{array}[]{cc}1&0\\\ 0&0\end{array}\right)$ and
$1-BA=\left(\begin{array}[]{cc}0&-1\\\ 0&1\end{array}\right)$. Clearly, $1-AB$
is a projection, and $1-BA$ is an idempotent but not a projection. Consulting
with Example 3.4(2), we readily see that $1-AB$ is $pns$-$*$-Drazin inverses,
while $1-BA$ is not.

$(2)$ An element $a$ of a $*$-ring is $pns$-$*$-Drazin invertible cannot imply
that so is $1-a$. For example, let $R=\mathbb{Z}_{6}$ be the ring of integers
modulo $6$. The involution of $R$ is defined by $*=1_{R}$, where $1_{R}$ is
the identity endomorphism of $R$. It is clear that $\sqrt{J(R)}=0$ and $0,1$
are the only projections of $R$. For $n=2\in\mathbb{N}$, due to Proposition
3.3, $a=-1\in R$ is $pns$-$*$-Drazin invertible, whereas $1-a=2\in R$ is
manifestly not.

We are now ready to prove the following assertion.

######  Theorem 3.11.

Let $R$ be a $*$-ring and $a$, $b$, $c\in R$. If $aba=ba^{2}=a^{2}c=aca$, then
$\alpha=1-ba\in R^{\diamond}$ if, and only if, $\beta=1-ac\in R^{\diamond}$.
In this case,

$\alpha^{\diamond}=1-b\beta^{\pi}v(ac)^{2}a+b\beta\beta^{\diamond}a+b\beta^{\diamond}aba$,

and

$\beta^{\diamond}=1-a\alpha^{\pi}u(ba)^{2}c+a\alpha\alpha^{\diamond}c+a\alpha^{\diamond}cac$,

where $\alpha^{\pi}=1-\alpha\alpha^{\diamond}$,
$\beta^{\pi}=1-\beta\beta^{\diamond}$,
$v=(1-\beta\beta^{\pi}\sum_{i=0}^{2}(ac)^{i})^{-1}$ and
$u=(1-\alpha\alpha^{\pi}\sum_{i=0}^{2}(ba)^{i})^{-1}$.

###### Proof.

Assume that $\alpha=1-ba$ is $pns$-$*$-Drazin invertible. Let

$u=(1-\alpha\alpha^{\pi}\sum_{i=0}^{2}(ba)^{i})^{-1}$

and

$x=1-a\alpha^{\pi}u(ba)^{2}c+a\alpha\alpha^{\diamond}c+a\alpha^{\diamond}cac.$

Consulting with Lemma 3.9, one concludes that $x\beta x=x$, $x\in{\rm
comm}^{2}(\beta)$ and $\beta^{n}-\beta x\in\sqrt{J(R)}$. To show that
$\beta\in R^{\diamond}$, it suffices to prove $(\beta x)^{\ast}=\beta x$.
However, in view of the proof of Lemma 3.9, it must be that
$\alpha^{\pi}=\alpha^{\pi}(ba)^{3}u$ and $\beta x=1-a\alpha^{\pi}u(ba)^{2}c$.
Since $a\in{\rm comm}(ba),$ the element $a$ commutes with both $\alpha^{\pi}u$
and $ba$. By hypothesis, we obtain

$\beta
x=1-a\alpha^{\pi}u(ba)^{2}c=1-\alpha^{\pi}u(ba)^{2}ac=1-\alpha^{\pi}u(ba)^{3}=1-\alpha^{\pi}=\alpha\alpha^{\diamond}.$

Thus, $(\beta x)^{\ast}=\beta x$, as desired. The converse can be shown
analogously. ∎

As an automatic consequence, we yield:

######  Corollary 3.12.

Let $R$ be a $*$-ring and $a$, $b\in R$. If $aba=ba^{2}=a^{2}b$, then
$\alpha=1-ba\in R^{\diamond}$ if, and only if, $\beta=1-ab\in R^{\diamond}$.

##  4\. $pns$-Drazin inverses for all $n\geq 1$

Here are focussing on the description of some crucial properties of rings for
which every element has a $pns$-Drazin inverse (the integer $n$ depending on
the choice of the element). We shall say that a ring $R$ is pseudo
$\pi$-polar, provided all of its elements are $pns$-Drazin invertible for
various naturals $n$; or, equivalently, for each $a\in R$, there exists
$e^{2}=e\in{\rm comm}^{2}(a)$ such that $a^{n}-e\in\sqrt{J(R)}$ for some
integer $n\geq 1$. Thus, this leads to the interesting fact that pseudo
$\pi$-polar rings are strongly clean in the sense that any of their elements
is a sum of a unit and an idempotent which commute.

On the other side, it was considered in [9] those rings $R$ in which every
$u\in U(R)$ satisfies the relation $u^{n}-1\in R^{\rm nil}$ for some $n\geq 1$
(which are called _$\pi$ -UU rings_). It was shown there that strongly
$\pi$-regular $\pi$-UU rings are always periodic in the sense that, for each
$a\in R$, there are two integers $m>n\geq 1$ with $a^{m}=a^{n}$. It is long
known that $J(R)$ is nil whenever $R$ is strongly $\pi$-regular. However, this
implication is still unproved for $\pi$-UU rings.

And so, our first goal here is to find a suitable connection between strongly
$\pi$-regular rings, $\pi$-UU rings and pseudo $\pi$-polar rings. That is, as
aforementioned, it is principally known that both strongly $\pi$-regular and
pseudo $\pi$-polar rings are strongly clean. In this vein, with the results
established above at hand, it is not too hard to prove that pseudo $\pi$-polar
rings $R$ with nil $J(R)$ are always periodic, and vice versa.

Concretely, the following is fulfilled.

######  Theorem 4.1.

A ring $R$ is periodic if, and only if, $R$ is pseudo $\pi$-polar and $J(R)$
is nil.

###### Proof.

Assume that $R$ is periodic. Then, $R$ is strongly $\pi$-regular. In view of
[24, Theorem 2.1], $R$ is pseudo-polar and $J(R)$ is nil. Furthermore, let
$u\in U(R)$. Then, there exist integers $k>l\geq 1$ such that $u^{k}=u^{l}$.
So, $1-u^{k-l}=0\in\sqrt{J(R)}$. Hence, $R$ is pseudo-$\pi$-polar by
exploiting Theorem 2.6, as claimed.

Conversely, letting $a\in R$, in view of Theorem 2.6 we write
$a-a^{n+1}\in\sqrt{J(R)}$ for some integer $n\geq 1$. Since $J(R)$ is nil, we
get $\sqrt{J(R)}=R^{\rm nil}$. Thus, $a-a^{n+1}$ in nilpotent, i.e.,

$(a-a^{n+1})^{l}=a^{l}(1-a^{n})^{l}=0$

for some positive integer $l$. This yields that $a^{l}=a^{l+1}f(a)$ for some
polynomial $f(x)\in\mathbb{Z}[x]$. Finally, by virtue of [1, Theorem 2.1], we
conclude that $R$ is a periodic ring, as asserted. ∎

Recall that an element $a$ of a ring $R$ is said to be _quasi-nilpotent_ [12]
if $1-ax\in U(R)$ for all $x\in{\rm comm}(a)$. The set of all quasi-nipotent
elements of $R$ is denoted by $R^{\rm qnil}$. We can infer from [24, Example
4.2] that $\sqrt{J(R)}$ is properly contained in $R^{\rm qnil}$.

Nevertheless, the following is true.

######  Proposition 4.2.

If $R$ is a pseudo $\pi$-polar ring, then $\sqrt{J(R)}=R^{\rm qnil}$.

###### Proof.

It suffices to show that $R^{\rm qnil}\subseteq\sqrt{J(R)}$. To that purpose,
let $q\in R^{\rm qnil}$. Then, there is an integer $n\geq 1$ such that
$q-q^{n+1}\in\sqrt{J(R)}$. Write $u=1-q^{n}$. Since $q\in R^{\rm qnil}$, we
get $u\in U(R)$ and $uq=qu\in\sqrt{J(R)}$. This implies
$q=(qu)u^{-1}\in\sqrt{J(R)}$. Therefore, $\sqrt{J(R)}=R^{\rm qnil}$, as
required. ∎

Thereby, combining the results quoted above, one verifies that the following
relationships are valid:

strongly $\pi$-regular + pseudo $\pi$-polar $\iff$ periodic $\iff$ strongly
$\pi$-regular + $\pi$-UU.

Taking into account the discussion alluded to above, a question which
automatically arises is what can be said about pseudo $\pi$-polar $\pi$-UU
rings, i.e., are they strongly $\pi$-regular and thus periodic? We note that
there is a pseudo $\pi$-polar ring which is not $\pi$-UU. Indeed, let
$R=\mathbb{Z}_{(p)}$ be the localization of the ring of integers $\mathbb{Z}$
at the prime ideal $(p)$, where $p$ is a prime integer. In virtue of [24,
Example 2.10], $R$ is pseudo-polar. Observe that $R/J(R)\cong\mathbb{Z}_{p}$.
Furthermore, for each $u\in U(\mathbb{Z}_{(p)})$, one has that
$\overline{u}^{p-1}=\overline{1}\in R/J(R)$. Thus, $1-u^{p-1}\in J(R)$.
Utilizing Theorem 2.6, $R$ is a pseudo-$\pi$-polar ring, but manifestly it is
not $\pi$-UU.

Specifically, we pose the following which, hopefully, makes some sense:

######  Conjecture 4.3.

Pseudo $\pi$-polar $\pi$-UU rings are strongly $\pi$-regular, and hence
periodic.

We just notice that, if $J(R)$ is nil, this will imply a positive resolution
of above problem.

We continue our considerations with some matrix results like these in the
sequel. We first need the following technicality.

######  Proposition 4.4.

If $R$ is a pseudo-$\pi$-polar ring, then so is the corner subring $eRe$ for
any $e^{2}=e\in R$.

###### Proof.

Let $a\in eRe$. One inspects that Theorem 2.6 allows us to derive that
$a-a^{n+1}\in\sqrt{J(R)}$ for some integer $n\geq 1$. So, there exists an
integer $k$ such that $(a-a^{n+1})^{k}\in J(R)$. Observe also that
$(a-a^{n+1})^{k}\in eRe$. Then,

$(a-a^{n+1})^{k}\in J(R)\cap eRe=J(eRe),$

that is, $a-a^{n+1}\in\sqrt{J(eRe)}$. Furthermore, as $R$ is pseudo-polar, in
accordance with [24, Corollary 2.10] we conclude that $eRe$ is pseudo-polar
too. Thus, the result follows now immediately. ∎

All local rings are known to be pseudo-polar (see cf. [24]). But,
interestingly, there exist local rings which are not pseudo $\pi$-polar, e.g.,
such as the rational number field $\mathbb{Q}$ and any field that contains
$\mathbb{Q}$. It is easy to see that a local ring $R$ is pseudo $\pi$-polar
if, and only if, for each $u\in U(R)$, $u^{n}\in 1+J(R)$ for some $n\geq 1$.
For a convenience in the exposition, we call a local ring $R$ _special_ if,
for every $u\in U(R)$, $u^{n}\in 1+J(R)$ for some $n\geq 1$. For example, the
considered above $\mathbb{Z}_{(p)}$ is a special local ring.

For a ring $R$, the $n\times n$ upper triangular matrix ring over $R$ is
denoted by ${\rm T}_{n}(R)$.

######  Proposition 4.5.

Let $R$ be a commutative local ring. The following are equivalent$:$

$(1)$ $R$ is a special local ring.

$(2)$ ${\rm T}_{n}(R)$ is pseudo $\pi$-polar for any integer $n\geq 1$.

$(3)$ ${\rm T}_{n}(R)$ is pseudo $\pi$-polar for some integer $n\geq 1$.

###### Proof.

$(1)\Rightarrow(2)$. Since $R$ is a commutative local ring, by using [24,
Theorem 2.13] ${\rm T}_{n}(R)$ is pseudo-polar. One verifies that, in view of
Theorem 2.6, we only need to show that, for every $V=(v_{ij})\in U({\rm
T}_{n}(R))$, there exists integer $k\geq 1$ such that $I-U^{k}\in J({\rm
T}_{n}(R))$. In fact, $V\in U({\rm T}_{n}(R))$ if, and only if, $v_{ii}\in
U(R)$ for each $i$. As $R$ is a special local ring, we check that
$1-v_{ii}^{k_{i}}\in J(R)$, where $k_{i}\geq 1$ and $i=1,\ldots,n$. Put
$k=k_{1}k_{2}\cdots k_{n}$. Thus,

$1-v_{ii}^{k}=(1-v_{ii}^{k_{i}})c\in J(R)$

for some $c\in R$. It follows immediately that $I-U^{k}\in J({\rm T}_{n}(R))$,
as needed. So, ${\rm T}_{n}(R)$ is pseudo $\pi$-polar, as promised.

$(2)\Rightarrow(3)$ is clear.

$(3)\Rightarrow(1)$. Taking into account Proposition 4.4, $R$ is pseudo
$\pi$-polar, and the rest is quite obvious. ∎

It can be inferred from [6] that matrix rings over a special local ring need
not be strongly clean. So, in particular, matrix rings over pseudo $\pi$-polar
rings are not always pseudo $\pi$-polar.

It was established in [9] that if $R$ is a ring such that $a^{m}=a$ for all
$a\in R$ and a fixed integer $m>1$, then, for any positive integer $n$, the
matrix ring ${\rm M}_{n}(R)$ is periodic. We now expand this to the following
affirmation.

######  Proposition 4.6.

Let $R$ be a commutative ring. If $a-a^{m}\in R^{\rm nil}$ for each $a\in R$
and a fixed integer $m>1$, then ${\rm M}_{n}(R)$ is pseudo $\pi$-polar for any
$n\geq 1$.

###### Proof.

In view of [16], $R$ is strongly $\pi$-regular and every element $a$ from
$\overline{R}=R/J(R)$ satisfies the identity $\overline{a}^{m}=\overline{a}$.
Since $R$ is a commutative strongly $\pi$-regular ring, it follows from [4]
that ${\rm M}_{n}(R)$ is strongly $\pi$-regular. Bearing in mind [24, Theorem
2.1], one concludes that ${\rm M}_{n}(R)$ is pseudo-polar.

Now, we finish the proof by showing that, for every $V\in U(M_{n}(R))$, the
relation $1-V^{k}\in J({\rm M}_{n}(R))$ holds for some integer $k$ or, in an
equivalent manner, that, for each element

$\overline{V}\in U(M_{n}(\overline{R}))=U(M_{n}(R)/J(M_{n}(R))),$

the equality $\overline{V}^{k}=\overline{1}$ holds for some integer $k$. So,
with no harm in generality, we may assume that $J(R)=0$. Hence, $a=a^{m}$ for
all $a\in R$. By the arguments of [17, p. 197], $R$ is a subdirect product of
its left primitive homomorphic images $R_{i}$ such that each $R_{i}$ is a
field satisfying $a_{i}^{m}=a_{i}$ for all $a_{i}\in R_{i}$ and $i\in\Lambda$.
Therefore, $|R_{i}|\leq m$ for all $R_{i}$, and thus it must be that $|{\rm
M}_{n}(R_{i})|\leq m^{n^{2}}$. Set $k=m^{n^{2}}!$. Then, for any $U_{i}\in
U({\rm M}_{n}(R_{i}))$, we can get that $U_{i}^{k}=1$. Observing that ${\rm
M}_{n}(R)$ can be viewed as a unitary subring of ${\rm
M}_{n}(\prod_{i\in\Lambda}R_{i})\cong\prod_{i\in\Lambda}{\rm M}_{n}(R_{i})$.
Let $V=(U_{i})_{i\in\Lambda}$ be a unit of $\prod_{i\in\Lambda}{\rm
M}_{n}(R_{i})$. Consequently, for each $i$, $U_{i}$ is a unit of ${\rm
M}_{n}(R_{i})$. From the equality $U_{i}^{k}=1$, we obtain that $V^{k}=1$, as
required. ∎

ACKNOWLEDGMENTS:

The research of the first-named and third-named authors was supported by the
Key Laboratory of Financial Mathematics of Fujian Province University (Putian
University) (No. JR202203), Anhui Provincial Natural Science Foundation (No.
2008085MA06) and the Key project of Anhui Education Committee (No.
gxyqZD2019009). The research of the second-named author was supported by the
projects Junta de Andalucía under Grant (FQM 264) and TÜBÍTAK under Grant
(BIDEB 2221).

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