On Generalized Rickart $*$-rings

TL;DR

This paper introduces and characterizes generalized Rickart *-rings, expanding the understanding of their structure through new concepts like generalized right projections and weakly Rickart *-rings.

Contribution

It defines generalized Rickart *-rings, provides their characterizations, and explores their properties including projections, weak variants, and orthogonal decompositions.

Findings

Every element has a generalized right projection.

Generalized Rickart *-rings satisfy the parallelogram law.

Projections in these rings have orthogonal decompositions.

Abstract

A ring $R$ with an involution $*$ is a generalized Rickart $*$-ring if for all $x\in R$ the right annihilator of $x^n$ is generated by a projection for some positive integer $n$ depending on $x$. In this work, we introduce generalized right projection of an element in a $*$-ring and prove that every element in a generalized Rickart $*$-ring has generalized right projection. Various characterizations of generalized Rickart $*$-rings are obtained. We introduce the concept of generalized weakly Rickart $*$-ring and provide a characterization of generalized Rickart $*$-rings in terms of weakly generalized Rickart $*$-rings. It is shown that generalized Rickart $*$-rings satisfy the parallelogram law. A sufficient condition is established for partial comparability in generalized Rickart $*$-rings. Furthermore, it is proved that pair of projections in a generalized Rickart $*$-ring possess orthogonal decomposition.

Equations

$$A_n(R)=R{I}_n+\sum _{l=2}^{[\frac{n}{2}]}R{V}_n^{l-1}+\sum _{i=1}^{[\frac{n+1}{2}]}\sum _{j=[\frac{n}{2}]+i}^{n}R{E}_{ij}.$$

$$B_n(R)=R{I}_n+\sum _{l=3}^{[\frac{n}{2}]}R{V}_n^{l-2}+\sum _{i=1}^{[\frac{n+1}{2}]+1}\sum _{j=[\frac{n}{2}]+i-1}^{n}R{E}_{ij}.$$

$$U_n(R)=R{I}_n+\sum _{i=1}^{[\frac{n-1}{2}]}\sum _{j=[\frac{n}{2}]+1}^{n}R{E}_{ij}+\sum _{j=[\frac{n-1}{2}]+2}^{n}R{E}_{[\frac{n-1}{2}]+1,j}.$$

$$S_n(R)=R{I}_n+\sum _{i<j}R{E}_{i,j},V_n(R)=R{I}_n+\sum _{l=2}^{n}R{V}_n^{l-1}.$$

Full text

Abstract: A ring $R$ with an involution $*$ is a generalized Rickart $*$-ring if for all $x\in R$ the right annihilator of $x^{n}$ is generated by a projection for some positive integer $n$ depending on $x$. In this work, we introduce generalized right projection of an element in a $*$-ring and prove that every element in a generalized Rickart $*$-ring has generalized right projection. Various characterizations of generalized Rickart $*$-rings are obtained. We introduce the concept of generalized weakly Rickart $*$-ring and provide a characterization of generalized Rickart $*$-rings in terms of weakly generalized Rickart $*$-rings. It is shown that generalized Rickart $*$-rings satisfy the parallelogram law. A sufficient condition is established for partial comparability in generalized Rickart $*$-rings. Furthermore, it is proved that pair of projections in a generalized Rickart $*$-ring possess orthogonal decomposition.

Keywords: generalized Rickart $*$-ring, generalized right projection, projections, generalized weakly Rickart $*$-ring.

1. Introduction

Kaplansky [5] introduced Baer rings and Baer $*$-rings to generalize various properties of $AW^{*}$-algebras (i.e., a $C^{*}$-algebra which is also a Baer $*$-ring), von Neumann algebras and complete $*$-regular rings. The concept of a Baer $*$-ring arises naturally from the study of functional analysis. For instance, every von Neumann algebra is a Baer $*$-algebra. For recent work on rings with involution, one can refer to [6, 7, 8, 9, 10, 11].

A ring $R$ is said to be reduced if it does not contains nonzero nilpotent element. A ring $R$ is said to be abelian if its every idempotent element is central. Let $S$ be a nonempty subset of $R$. We write $r(S)=\{a\in R~:~sa=0,~\forall~s\in S\}$, and is called the right annihilator of $S$ in $R$, and $l(S)=\{a\in R~|~as=0,~\forall~s\in S\}$, is the left annihilator of $S$ in $R$. Let $R$ be a ring and $a\in R$, then we write $r(\{a\})=r(a)$ and $l(\{a\})=l(a)$. A $*$-ring $R$ is a ring equipped with an involution $x\rightarrow x^{*}$, that is, an additive anti-automorphism of the period at most two. An element $e$ of a $*$-ring $R$ is called a projection if it is self-adjoint (i.e. $e=e^{*}$) and idempotent (i.e. $e^{2}=e$). Let $P$ be a poset and $a,b\in P$. The join of $a$ and $b$, denoted by $a\vee b$, is defined as $a\vee b=\sup~\{a,b\}$. The meet of $a$ and $b$, denoted by $a\wedge b$, is defined as $a\wedge b=inf~\{a,b\}$. In a poset $(P,\leq)$, $a<b$ denotes $a\leq b$ with $a\neq b$. Let $R$ be a $*$-ring and $e,f\in R$ be projections, we say that $e\leq f$ if $e=ef$, this defines a partial order on the set of all projections in $R$. A $*$-ring $R$ is said to be a Rickart $*$-ring, if for each $x\in R$, $r(x)=eR$, where $e$ is a projection in $R$. For each element $x$ in a Rickart $*$-ring, there is unique projection $e$ such that $xe=x$ and $xy=0$ if and only if $ey=0$, called the right projection of $x$, denoted by $RP(x)$. Similarly, the left projection $LP(x)$ is defined for each element $x$ in Rickart $*$-ring. A $*$-ring $R$ is said to be a weakly Rickart $*$-ring, if for any $x\in R$, there exists a projection $e$ such that (1) $xe=x$, and (2) if for $y\in R$, $xy=0$ then $ey=0$.

In [3], Berberian posed the following open problem.

Problem 1: Can every weakly Rickart $*$-ring be embedded in a Rickart $*$-ring with preservation of $RP$’s?

In [3] Berberian provided a partial solution to this problem. Subsequently, in [12], authors offered another partial solution. In [10], a more general partial solution to Problem 1 is presented.

Let $R$ be a $*$-ring. The projections $e,f$ in $R$ are said to be equivalent (written as $e\sim f$), if there exists $w\in R$ such that $w^{*}w=e$ and $ww^{*}=f$ (see [3]). By [3, Proposition 7, page 5], the relation $\sim$ is an equivalence relation on the set of projections in $R$. A $*$- ring is said to satisfy parallelogram law if $e-e\wedge f\sim e\vee f-f$ for every pair of projection $e$ and $f$ whose $e\wedge f$ and $e\vee f$ exists. Projections $e$ and $f$ in a $*$-ring $R$ are said to be partially comparable if there exist non-zero projections $e_{0}$ and $f_{0}$ such that $e_{0}\leq e$, $f_{0}\leq f$ and $e_{0}\sim f_{0}$. We say that a $*$-ring $R$ has partial comparability $(PC)$ if $eRf\neq 0$ implies $e$ and $f$ are partially comparable. Let $R$ be a $*$-ring and $e,f$ be projections in $R$. We say that $e$ is *dominated * by $f$, written as $e\lesssim f$, if $e\sim g\leq f$ for some projection $g\in R$. Projections $e$ and $f$ in a $*$-ring $R$ are said to be generalized comparable if there exists central projection $h$ such that $he\lesssim hf$ and $(1-h)f\lesssim(1-h)e$. We say that $R$ has generalized comparability $(GC)$ if every pair of projection in $R$ is generalized comparable. We say that a $*$-ring $R$ has orthogonal GC if every pair of orthogonal projections are generalized comparable. Projections $e$ and $f$ in a $*$-ring $R$ are said to be very orthogonal if there exists central projection $h$ such that $he=e$ and $hf=0$.

In [4], the authors introduced the concept of generalized Rickart $*$-ring. A $*$-ring $R$ is called a generalized Rickart $*$-ring, if, for any $x\in R$, there exists a positive integer $n$ such that $r(x^{n})=gR$, for some projection $g$ of $R$. In generalized Rickart $*$-rings, we also have $l(x^{n})=Rh$ for some projection $h\in R$. This indicates that generalized Rickart $*$-rings are left-right symmetric. Generalized Rickart $*$-rings serve as a common generalization of both Rickart $*$-rings and generalized Baer $*$-rings. In [2], M. Ahmadi and A. Moussavi explored the behavior of the generalized Rickart $*$-condition under various constructions and extensions. They also provided examples of generalized Rickart $*$-rings and identified classes of finite and infinite-dimensional Banach $*$-algebras that are generalized Rickart $*$-rings but not Rickart $*$-rings.

In the second section of this paper, we introduce generalized right projection of an element in a $*$-ring and prove that every element of a generalized Rickart $*$-ring has a generalized right projection. Properties of generalized right projection of elements in a generalized Rickart $*$-ring are also studied. We introduce the concept of a generalized weakly Rickart $*$-ring and provide a characterization of generalized Rickart $*$-rings in terms of generalized weakly Rickart $*$-rings. It is shown that the center and corner of a generalized weakly Rickart $*$-ring are themselves generalized weakly Rickart $*$-ring. In section 3, we pose a problem for generalized Rickart $*$-rings analogous to Problem 1 for Rickart $*$-rings and provide a partial solution. In section 4, we prove that generalized Rickart $*$-rings satisfy the parallelogram law. A sufficient condition is established for generalized Rickart $*$-rings to exhibit partial comparability. Furthermore, it is shown that pairs of projections in a generalized Rickart $*$-ring that satisfying the parallelogram law possess orthogonal decomposition.

2. Weakly Generalized Rickart $*$-rings

It is evident that every Rickart $*$-ring is a generalized Rickart $*$-ring. In this section, we first recall examples and results from [2], which provides instances of generalized Rickart $*$-rings that are not Rickart $*$-rings. We then introduce generalized right projection of an element and generalized left projection of an element in a $*$-ring. We prove that every element of a generalized Rickart$*$-ring has a generalized right projection and discuss the properties of these projections. Furthermore, we introduce the class of generalized weakly Rickart $*$-rings and establish characterizations of generalized Rickart $*$-rings.

Example 2.1** ([2], Example 2.8).**

(i) Let $S=C[x,y]/(x,y)^{n}$. Then $S$ is a commutative local ring with unique maximal ideal $(\overline{x},\overline{y})$ having index of nilpotency $n$. The set $S$ with the conjugate map as the involution, is a generalized Rickart $*$-ring but not Rickart $*$-ring.

(ii) Let $G$ be a finite abelian $p$-group. Then the group algebra $T=F_{p}G$ is a finite commutative local ring with unique maximal ideal $rad(T)$ having index of nilpotency $|G|$. Let $*$ be the involution on the group ring $T$ defined by $(\sum a_{g}g)^{*}=\sum a_{g}g^{-1}$. Hence $T$ is a generalized Rickart $*$-ring but not Rickart $*$-ring.

(iii) Let $S$ and $T$ be the rings in (i) and (ii) respectively. Let $n$ be a positive integer and let $p$ be a prime. Then the $*$-ring $R=S\oplus T\oplus\mathbb{Z}_{p^{n}}$ is a generalized Rickart $*$-ring that is not Rickart $*$-ring.

(iv) Let $T=\{a_{0}+a_{1}i+a_{2}j+a_{3}k~:~a_{i}\in\mathbb{Z}_{2}~\text{for}~i=1,2,3,4\}$ be the Hamilton quaternions over $\mathbb{Z}_{2}$. Then $T$ is a commutative ring such that every element of $T$ is either invertible or nilpotent. For $\alpha=a_{0}+a_{1}i+a_{2}j+a_{3}k$ with $a_{0},a_{1},a_{2},a_{3}\in\mathbb{Z}_{2}$, define $\alpha^{*}=a_{0}-a_{1}i-a_{2}j-a_{3}k$. Then $*$ is an involution for $T$. Then $T$ is generalized Rickart $*$-ring but not a Rickart $*$ ring.

Definition 2.2**.**

([1], Definition 5.1). Let $R$ be a ring with unity, and let $n\geq 2$ be an integer. Put $V_{n}=\sum_{i=1}^{n-1}E_{i,i+1}$. The triangular matrix rings $S_{n}(R),A_{n}(R),B_{n}(R),U_{n}(R)$, and $V_{n}(R)$ are define as follows:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Proposition 2.3** ([2], Theorem 4.6).**

Let $A$ be an abelian $C^{*}$-algebra. If $A$ is a generalized Rickart $*$-ring, then the Banach $*$-algebras $S_{n}(\mathcal{A}),A_{n}(\mathcal{A}),B_{n}(\mathcal{A}),U_{n}(\mathcal{A})$, and $V_{n}(\mathcal{A})$ are generalized Rickart $*$-rings but not Rickart $*$-rings. In particular, the Banach $*$-algebras $S_{n}(C),A_{n}(C),B_{n}(C),U_{n}(C)$, and $V_{n}(C)$ are generalized Rickart $*$-rings but not Rickart $*$-rings.

Example 2.4** ([2], Example 4.7).**

Let $A$ be an infinite-dimensional commutative von Neumann algebra. Then $A$ is a Baer $*$-ring (and hence a Rickart $*$-ring). Now let $R$ be the ring $Sn(C)(An(C),Bn(C),Un(C),orVn(C))$ and put $T=A\oplus R$ (equipped with the max norm). Then $T$ is an infinite-dimensional Banach $*$-algebra and $T$ is a generalized Rickart $*$-ring but not a Rickart $*$-ring.

Following is an example of a finite generalized Rickart $*$-ring that is not a Rickart $*$-ring.

Example 2.5**.**

Let $R=\mathbb{Z}_{4}$ be a $*$-ring with an identity involution . Here $0,1$ are only projections in $\mathbb{Z}_{4}$. Now $r(2)=\{0,2\}\neq 0\mathbb{Z}_{4}$ and $r(2)=\{0,2\}\neq 1\mathbb{Z}_{4}$. Hence $r(2)$ is not generated by any projection in $R$. Therefore $\mathbb{Z}_{4}$ is not a Rickart $*$-ring. Observe that $r(0)=1\mathbb{Z}_{4},r(1)=0\mathbb{Z}_{4}$, $r(2^{2})=r(0)=1\mathbb{Z}_{4}$, and $r(3)=0\mathbb{Z}_{4}$. Hence for every $x\in\mathbb{Z}_{4},\exists~n\in\mathbb{N}$ and projection $e$ in $\mathbb{Z}_{4}$ such that $r(x^{n})=e\mathbb{Z}_{4}$. Therefore $\mathbb{Z}_{4}$ is a generalized Rickart $*$-ring.

The following is an example of a finite $*$-ring that is not a generalized Rickart $*$-ring.

Example 2.6**.**

Let $R=M_{4}(\mathbb{Z}_{4})$ with transpose as an involution. Then $R$ is not a generalized Rickart $*$-ring (see Example 2.13).

In the following result, we prove that generalized Rickart $*$-rings, always contain the multiplicative identity (unity) element.

Proposition 2.7**.**

If $R$ is a generalized Rickart $*$-ring then it has unity element.

Proof.

Let $R$ be a generalized Rickart $*$-ring. Since for any $n\in\mathbb{N}$, $r(0^{n})=r(0)=R=eR$ for some projection $e\in R$. Therefore for $x\in R$, we have $x=ey$ for some $y\in R$. This implies $ex=e^{2}y=ey=x$. Thus $e$ is the unity element in $R$. ∎

Proposition 2.8**.**

If $R$ is a generalized Rickart $*$-ring then for every $x\in R$ there exists $n\in\mathbb{N}$ and a projection $e\in R$ such that $x^{n}e=x^{n}$; and for $y\in R$, if $x^{n}y=0$ then $ey=0$.

Proof.

As $R$ is a generalized Rickart $*$-ring, for every $x\in R,r(x^{n})=gR$ for some $n\in\mathbb{N}$ and for some projection $g\in R$. Let $e=1-g$, then $x^{n}e=x^{n}(1-g)=x^{n}-x^{n}g=x^{n}$. Let $y\in R$ and $x^{n}y=0$. Then $y\in r(x^{n})=gR$, this gives $y=gy$, that is $(1-g)y=0$. Therefore $ey=0$. Thus $x^{n}y=0$ implies that $ey=0$. ∎

Observe that in the above proposition, the projection $e$ is the smallest projection such that $x^{n}e=x^{n}$ (see Proposition 2.14). Now, we introduce the generalized right projection of an element in a $*$-ring.

Definition 2.9**.**

Let $R$ be a $*$-ring and $x\in R$. The projection $e\in R$ is said to be generalized right projection of $x$ denoted by $GRP(x)$ if there exists $n\in\mathbb{N}$ such that $x^{n}e=x^{n}$; and for $y\in R$ if $x^{n}y=0$ then $ey=0$.

Similarly, we introduce the generalized left projection of an element in a $*$-ring.

Definition 2.10**.**

Let $R$ be a $*$-ring and $x\in R$. The projection $f\in R$ is said to be generalized left projection of $x$ denoted by $GLP(x)$ if there exists $n\in\mathbb{N}$ such that $fx^{n}=x^{n}$; and for $y\in R$ if $yx^{n}=0$ then $yf=0$.

Remark 2.11**.**

By Proposition 2.8, every element of a generalized Rickart $*$-ring possesses a generalized right projection. Similarly, every element of a generalized Rickart $*$-ring possesses a generalized left projection.

Definition 2.12**.**

A $*$-ring $R$ is said to be generalized weakly Rickart $*$-ring if every $x\in R$ possesses a generalized right projection. That is $GRP(x)$ exists for every $x\in R$.

By Remark 2.11, every generalized Rickart $*$-ring is a generalized weakly Rickart $*$-ring. The following is an example of a $*$-ring which is not a weakly generalized Rickart $*$-ring.

Example 2.13**.**

Let $R=M_{4}(\mathbb{Z}_{4})$ and $A=\begin{bmatrix}1&1&1&1&\\ 0&0&0&0&\\ 0&0&0&0&\\ 0&0&0&0&\end{bmatrix}$, $B=\begin{bmatrix}1&1&1&1&\\ 1&1&1&1&\\ 1&1&1&1&\\ 1&1&1&1&\\ \end{bmatrix}\in R$.

Suppose $GRP(A)=E$. Then there exits $n\in\mathbb{N}$, such that $A^{n}E=A^{n}$; and $A^{n}B=0$ implies $EB=0$. That is $AE=A$; and $AB=0$ implies $EB=0$. But $AE=A$ implies $e_{11}+e_{21}+e_{31}+e_{41}=1$. This gives $e_{11}+e_{12}+e_{13}+e_{14}=1$, a contradiction. Thus $M_{4}(\mathbb{Z}_{4})$ is not a generalized weakly Rickart $*$-ring.

Proposition 2.14**.**

Let $R$ be a generalized Rickart $*$-ring.

1. (1)

For $x,y\in R$, if $xy=0$ then $GRP(x)GLP(y)=0$. 2. (2)

If $GRP(x)=e$ then $e$ is the smallest projection such that $x^{n}e=x^{n}$ for some $n\in\mathbb{N}$.

Proof.

(1): Let $x,y\in R$ and $GRP(x)=e,~GLP(y)=f$. Then for some $n\in\mathbb{N}$, $x^{n}e=x^{n}$; and $x^{n}z=0$ implies $ez=0$. Also, for some $m\in\mathbb{N}$, $fy^{m}=y^{m}$; and $zy^{m}=0$ implies $zf=0$. Suppose $xy=0$. Then $x^{n-1}xy=0$ implies $x^{n}y=0$. This gives $ey=0$. Therefore $eyy^{m-1}=0$ implies $ey^{m}=0$. Hence $ef=0$. That is $GRP(x)GLP(y)=0$.

(2): Let $f$ be a projection such that $x^{n}f=x^{n}$. Then $x^{n}(e-f)=x^{n}e-x^{n}f=0$. This gives $e(e-f)=0$, that is $e=ef$. Hence $e\leq f$. Therefore $e$ is the smallest projection such that $x^{n}e=x^{n}$. ∎

The converse of Proposition 2.14 (1) is not true.

Example 2.15**.**

Let $R=\mathbb{Z}_{12}$. Observe that $0,1,4,9$ are the only projections in $R$. Let $x=2,~y=3\in R$. Since $2^{2}\cdot 4=4=2^{2}$, we have $GRP(2)=4$. Also, $9\cdot 3^{2}=9=3^{2}$, gives $GLP(3)=9$. Therefore $GRP(2)GLP(3)=4\cdot 9=0$, but $2\cdot 3=6\neq 0$. Thus $GRP(x)~GLP(y)=0$ but $xy\neq 0$.

In the following result, we provide a condition under which a $*$-subring of a generalized Rickart $*$-ring becomes a generalized Rickart $*$-ring.

Proposition 2.16**.**

Let $R$ be a generalized Rickart $*$-ring and $B$ be a $*$-subring of $R$ satisfying the following conditions,

1. (1)

$B$* has unity element* 2. (2)

if $x\in B$ then $GRP(x)\in B$.

Then $B$ is a generalized Rickart $*$-ring.

Proof.

Let $x\in B$ and $GRP(x)=e$. Therefore for some $n\in\mathbb{N}$, $x^{n}e=x^{n}$; and $x^{n}y=0$ implies $ey=0$. Since $e\in B$. for $y\in B,~x^{n}y=0$ if and only if $ey=0$ if and only if $y=y-ey$ if and only if $y=(1-e)y$. Therefore $r(x^{n})=(1-e)B$. Hence $B$ is a generalized Rickart $*$-ring. ∎

Let $R$ be a ring and $S$ be a nonempty subset of $R$, the commutant of $S$ in $R$, denoted $S^{\prime}$, is the set of elements of $R$ that commute with every element of $S$, that is $S^{\prime}=\{x\in R~:~xs=sx,$ for all $s\in S\}$. We write $S^{\prime\prime}=(S^{\prime})^{\prime}$, called the bicommutant of $S$.

In the following result, we provide a condition under which a $*$-subring of a generalized Rickart $*$-ring becomes a generalized Rickart $*$-ring, and generalized right projection of every element in the $*$-subring remains within it.

Proposition 2.17**.**

Let $R$ be a generalized Rickart $*$-ring and $B$ be a $*$-subring of $R$ such that $B=B^{\prime\prime}$. Then

1. (1)

$x\in B$* implies $GRP(x)\in B$.* 2. (2)

$B$* is a generalized Rickart $*$-ring.*

Proof.

(1): Suppose $x\in B$ thus $x\in R$. Let $GRP(x)=e$. Therefore there exists $n\in\mathbb{N}$ such that $x^{n}e=x^{n}$; and for $y\in R$ $x^{n}y=0$ implies $ey=0$. To prove $e\in B=B^{\prime\prime}=(B^{\prime})^{\prime}$, it is enough to show $ey=ye$ for all $y\in B^{\prime}$. Now $xy=yx$ gives $x^{n}y=yx^{n}$. Therefore $x^{n}(y-ye)=x^{n}y-x^{n}ye=x^{n}y-yx^{n}e=x^{n}y-yx^{n}=0$. This implies $e(y-ye)=0$ and hence $ey=eye$. Replace $y$ by $y^{*}$. So $ey^{*}=ey^{*}e$, that is $ye=eye=ey$. Therefore $ey=ye$ for all $y\in B^{\prime}$. Hence $e\in B$, that is $GRP(x)\in B$.

(2): We know for any nonempty subset $S$ of a ring $R$, $1\in S^{\prime}$ and $S^{\prime}=S^{\prime\prime\prime}$. Since $B=B^{\prime\prime}$ is equivalent to $B=S^{\prime}$ for some $*$-subset $S$ of $R$, we have $1\in B$. By (1), $x\in B$ implies $GRP(x)\in B$. By Proposition 2.16, $B$ is a generalized Rickart $*$-ring. ∎

Proposition 2.18**.**

Let $R$ be a generalized Rickart $*$-ring and $x\in R$. Then there exists $n\in\mathbb{N}$ such that $r(x^{n})=(1-GRP(x))R$.

Proof.

Let $x\in R$ and $GRP(x)=e$. Therefore there exist $n\in\mathbb{N}$ such that $x^{n}e=x^{n}$; and for $z\in R$, $x^{n}z=0$ implies $ez=0$. We prove that $r(x^{n})=(1-e)R$. Let $y\in r(x^{n})$, then $x^{n}y=0$, and hence $ey=0$. Therefore $y=y-ey=(1-e)y\in(1-e)R$. Hence $r(x^{n})\subseteq(1-e)R$. Let $w\in(1-e)R$, then $w=(1-e)w=w-ew$. Hence $ew=0$ implies $x^{n}ew=0$. Which gives $x^{n}w=0$, and hence $w\in r(x^{n})$. Therefore $(1-e)R\subseteq r(x^{n})$. Thus $r(x^{n})=(1-e)R=(1-GRP(x))R$. ∎

In the following result, we provide a characterization of a generalized Rickart $*$-ring

Proposition 2.19**.**

A $*$-ring $R$ is generalized Rickart $*$-ring if and only if $R$ has unity element and for each $x\in R$ there exists a projection $e$ such that $r(x^{n})=r(e)$ for some $n\in\mathbb{N}$.

Proof.

Suppose $R$ is a generalized Rickart $*$-ring. Let $x\in R$ and $GRP(x)=e$. Therefore for some $n\in\mathbb{N}$, $x^{n}e=x^{n}$; and $x^{n}y=0$ if and only if $ey=0$. Hence $y\in r(x^{n})$ if and only if $y\in r(e)$ for some $n\in\mathbb{N}$. Thus $r(x^{n})=r(e)$. Conversely, suppose $R$ has unity element and for each $x\in R$ there exists a projection $e$ such that $r(x^{n})=r(e)$ for some $n\in\mathbb{N}$. Therefore $r(x^{n})=r(e)=(1-e)R$ for some $n\in\mathbb{N}$. Therefore for any $x\in R$, there exists a projection $1-e$ such that $r(x^{n})=(1-e)R$ for some $n\in\mathbb{N}$. Hence $R$ is a generalized Rickart $*$-ring. ∎

Following is a characterization of a generalized Rickart $*$-ring in terms of generalized weakly Rickart $*$-rings.

Proposition 2.20**.**

The following conditions on a $*$-ring $R$ are equivalent.

1. (1)

$R$* is a generalized Rickart $*$-ring.* 2. (2)

$R$* is a generalized weakly Rickart $*$-ring with unity.*

Proof.

Suppose $R$ is a generalized Rickart $*$-ring. By Proposition 2.7, $R$ has unity element. Also, by Proposition 2.8, $R$ is a generalized weakly Rickart $*$-ring. Therefore $R$ is a generalized weakly Rickart $*$-ring with unity. Conversely suppose $R$ is a generalized weakly Rickart $*$-ring with unity. Let $x\in R$. Then $GRP(x)$ exists in $R$. Therefore $r(x^{n})=(1-GRP(x))R$ for some $n\in\mathbb{N}$. Hence $R$ is a generalized Rickart $*$-ring. ∎

Following is the example of weakly generalized Rickart $*$-ring which is not generalized Rickart $*$-ring.

Example 2.21**.**

Let $R=\{x=(x_{1},x_{2},\cdots)~|~x_{i}\in\mathbb{C},~for~each~x,~there~exists~m\in\mathbb{N}~such~that~x_{k}=0~for~all~k>m\}$. Then $R$ is a ring with component-wise operations multiplication and addition. Define the involution $*$ on $R$ as, for $x=(x_{1},x_{2},\cdots)\in R,~x^{*}=(\bar{x_{1}},\bar{x_{2}},\cdots)$. Clearly $e=(e_{1},e_{2},\cdots)\in R$ is a projection if $e_{i}\in\{0,1\}$. Also, the unity element is $u=(1,1,\cdots)$ and $u\notin R$ . Let $x=(x_{1},x_{2},\cdots)\in R$. Then there exists $m\in\mathbb{N}$ such that $x_{k}=0$ for all $k>m$. Let us find $n\in\mathbb{N}$ and projection $e\in R$ such that $x^{n}e=x^{n}$ and $x^{n}y=0$ implies $ey=0$. Choose $n=1$ and define $e=(e_{1},e_{2},\cdots)$ as $e_{i}=1$ if $x_{i}\neq 0$ and $e_{i}=0$ if $x_{i}=0$. As $x$ has only finitely many non-zero components, $e$ also has finitely many non-zero components. So $e\in R$. Further, $e_{i}\in\{0,1\}$ and $e_{i}=\bar{e_{i}}$, thus $e$ is a projection in $R$. For each component $(xe)_{i}=x_{i}e_{i}$. If $x_{i}\neq 0$ then $e_{i}=1$, so $(xe)_{i}=x_{i}\cdot 1=x_{i}$. If $x_{i}=0$ then $e_{i}=0$, so $(xe)_{i}=0=x_{i}$. Thus, $xe=x$. Now suppose $xy=0$. This means $x_{i}y_{i}=0$ for all $i$. If $x_{i}\neq 0$ then $y_{i}$ must be 0. If $x_{i}=0$ then $y_{i}$ can be anything. We have $(ey)_{i}=e_{i}y_{i}$. Thus, If $x_{i}\neq 0$ then $e_{i}=1$. so $(ey)_{i}=1\cdot 0=0$. If $x_{i}=0$ then $e_{i}=0$ and hence $(ey)_{i}=0$. Therefore, $ey=0$. Thus $R$ is a weakly generalized Rickart $*$-ring. To show $R$ is not a generalized Rickart $*$-ring. We will find $x\in R$ such that for any $n\in\mathbb{N},~r(x^{n})\neq eR$ for any projection $e\in R$. Let $x=(0,1,0,0,\cdots)\in R$. We have $x^{n}=x$ for any $n\geq 1$. Let us find $r(x)$. Suppose $y=(y_{1},y_{2},\cdots)\in r(x)$. Hence, $xy=0$. Therefore $(0,1,0,0,\cdots)\cdot(y_{1},y_{2},\cdots)=(0,0,\cdots)$. This gives $y_{2}=0$. Thus, $r(x)=\{(y_{1},0,y_{3},y_{4},\cdots)~|~y_{i}\in\mathbb{C}\}$. Suppose $r(x)=eR$ for some projection $e\in R$. Let $e_{k}=0$ for $k>n_{0}$. If $y\in r(x)$, then $y\in eR$, thus $y_{i}=e_{i}z_{i}$ for some $z=(z_{1},z_{2},\cdots)\in R$. As $y_{2}=0$ for any $y\in r(x)$. So $(ez)_{2}=e_{2}z_{2}=0$. Hence $e_{2}=0$. Also, for $w=(1,0,0,\cdots)\in r(x)$, we have $e_{1}=1$. Similarly, $e_{3}=1$. Therefore $e=(1,0,1,1,1,\cdots)$. This contradicts the fact that $e$ has only finitely many non-zero components. Hence, $r(x^{n})\neq eR$ for any projection $e\in R$ and for any $n\in\mathbb{N}$. Thus, $R$ is not a generalized Rickart $*$-ring.

In the following result, we prove that the center of a generalized weakly Rickart $*$-ring is itself a generalized weakly Rickart $*$-ring.

Proposition 2.22**.**

The center of a generalized weakly Rickart $*$-ring is generalized weakly Rickart $*$-ring.

Proof.

Suppose $R$ is a generalized weakly Rickart $*$-ring. Let $C(R)$ denote the center of $R$ and $x\in C(R)$. We will prove that $GRP(x)$ exists in $C(R)$. Since $x\in R$ and $R$ is a generalized weakly Rickart $*$-ring. Therefore $GRP(x)=e$ exists in $R$. That is there exist $n\in\mathbb{N}$ such that $x^{n}e=x^{n}$; and $x^{n}y=0$ implies $ey=0$. Hence there exist $n\in\mathbb{N}$ such that $e(x^{n})^{*}=(x^{n})^{*}$; and $y^{*}(x^{n})^{*}=0$ implies $y^{*}e=0$. Since $x\in c(R)$, we have $x^{n}(r-re)=x^{n}r-x^{n}re=x^{n}r-rx^{n}e=x^{n}r-rx^{n}=x^{n}r-x^{n}r=0$. Therefore $e(r-re)=0$, that is $er-ere=0$, which gives $er=ere$. Also, $(r-er)(x^{n})^{*}=r(x^{n})^{*}-er(x^{n})^{*}=r(x^{n})^{*}-e(x^{n})^{*}r=r(x^{n})^{*}-(x^{n})^{*}r=r(x^{n})^{*}-r(x^{n})^{*}=0$. Hence $(r-er)e=0$ implies $re=ere$. Therefore $er=re$ for all $r\in R$. Hence $e\in C(R)$, that is $GRP(x)=e\in C(R)$. Thus $C(R)$ is a generalized weakly Rickart $*$-ring. ∎

The involution $*$ of a ring $R$ is called weakly proper if for any $x\in R$, $xx^{*}=0$ implies $x^{n}=0$ for some $n\in\mathbb{N}$.

Proposition 2.23**.**

*Let $R$ be a generalized weakly Rickart $*$-ring. Then,

1. (1)

for each $x\in R$ there exist $n\in\mathbb{N}$ such that $r(x^{n})\cap(x^{*})^{n}R=\{0\}$. 2. (2)

the involution $*$ is weakly proper.

Proof.

(1): Let $x\in R$. Since $R$ is a generalized weakly Rickart $*$-ring, there exists $n\in\mathbb{N}$ and a projection $e\in R$ such that $x^{n}e=x^{n}$; and for $y\in R$, $x^{n}y=0$ implies $ey=0$. We prove that $r(x^{n})\cap(x^{*})^{n}R=\{0\}$. Let $y\in r(x^{n})\cap(x^{*})^{n}R$. Therefore $x^{n}y=0$ and $y=(x^{*})^{n}s$ for some $s\in R$. This gives $y=(x^{*})^{n}s=(x^{n})^{*}s=(x^{n}e)^{*}s=e(x^{n})^{*}s=e(x^{*})^{n}s=ey=0$. Hence $r(x^{n})\cap(x^{*})^{n}R=\{0\}$.

(2): Let $xx^{*}=0$. Therefore $x^{n}(x^{*})^{n}=0$. This gives $e(x^{*})^{n}=0$. Hence $e(x^{n})^{*}=0$ implies $(x^{n}e)^{*}=0$, and this gives $(x^{n})^{*}=0$. Therefore $x^{n}=0$. ∎

Corollary 2.24** ([2], Proposition 2.11).**

*Let $R$ be a generalized Rickart $*$-ring. Then

(i) for each $x\in R$, there exists an integer $n\geq 1$ such that $r(x^{n})\cap(x^{*})^{n}R=0$; (ii) the involution $*$ is weakly proper.*

The following result provides the characterization of a generalized weakly Rickart $*$-ring.

Proposition 2.25**.**

The following conditions on a $*$-ring $R$ are equivalent.

1. (1)

$R$* is generalized weakly Rickart $*$-ring.* 2. (2)

Involution $*$ is weakly proper and for every $x\in R$ there exist $n\in\mathbb{N}$ and a projection $e$ in $R$ such that $r(x^{n})=r(e)$.

Proof.

Suppose $R$ is a generalized weakly Rickart $*$-ring. By Proposition 2.23, involution on $R$ is weakly proper. Let $x\in R$ and $GRP(x)=e$. Let $y\in r(x^{n})$. Therefore $x^{n}y=0$ implies $ey=0$, and hence $y\in r(e)$. Thus $r(x^{n})\subseteq r(e)$. Now let $y\in r(e)$. Therefore $ey=0$, which implies $x^{n}y=x^{n}ey=x^{n}0=0$. Hence $y\in r(x^{n})$. Therefore $r(e)\subseteq r(x^{n})$. Thus $r(x^{n})=r(e)$. Conversely, suppose involution $*$ is weakly proper and for every $x\in R$ there exist $n\in\mathbb{N}$ and a projection $e$ in $R$ such that $r(x^{n})=r(e)$. Let $x\in R$. Since $r(x^{n})=r(e)$, we have $e(1-e)=0$ implies $1-e\in r(x^{n})$. Therefore $x^{n}(1-e)=0$ implies $x^{n}=x^{n}e$. If $x^{n}y=0$ then $y\in r(x^{n})$. This gives $y\in r(e)$ and hence $ey=0$. Therefore $e=GRP(x)$ and $R$ is a generalized weakly Rickart $*$-ring. ∎

In the following result, we prove that the corner of a generalized weakly Rickart $*$-ring is itself a generalized weakly Rickart $*$-ring.

Proposition 2.26**.**

Let $R$ be a $*$-ring and $e$ be a projection in $R$. If $R$ is a generalized weakly Rickart $*$-ring then so is $eRe$.

Proof.

Let $x\in eRe$. Then $x\in R$. Since $R$ is a generalized weakly Rickart $*$-ring, $GRP(x)=f$ exists in $R$. Therefore there exist $n\in\mathbb{N}$ such that $x^{n}f=x^{n}$; and for $y\in R$, $x^{n}y=0$ implies $fy=0$. Since $x\in eRe$, we have $x=exe$. So $x^{n}e=(exe)^{n}e=(exe)^{n}=x^{n}$.Therefore $x^{n}(e-f)=x^{n}e-x^{n}f=x^{n}e-x^{n}$. Hence $x^{n}(e-f)=0$, which implies $f(e-f)=0$ thus $fe=f$. We prove that $GRP(x)=f$ in $eRe$. As above $x^{n}f=x^{n}$. Suppose $x^{n}(eze)=0$, then $feze=0$. Note that $f=ef=efe\in eRe$. Therefore $GRP(x)=f$ in $eRe$. Thus $eRe$ is a generalized weakly Rickart $*$-ring. ∎

Proposition 2.27**.**

Let R be a generalized weakly Rickart $*$-ring and $S$ be a self-adjoint subset of $R$ and $x\in S^{\prime}$. If $GRP(x)=e$ then $se=ese=es$ for all $s\in S$.

Proof.

Since $x\in S^{\prime}$, we have $xs=sx$ for all $s\in S$. As $GRP(x)=e$. Then there exists $n\in\mathbb{N}$ such that $x^{n}e=x^{n}$; and for $y\in R$, $x^{n}y=0$ implies $ey=0$. We have $x^{n}s=sx^{n}$. Now $x^{n}(se-es)=x^{n}se-x^{n}es=sx^{n}e-x^{n}s=sx^{n}-x^{n}s=0$. Therefore $e(se-es)=0$ implies $ese=es$. Replacing $s$ by $s^{*}$ we get $es^{*}e=es^{*}$. Therefore $ese=se$. Thus $se=ese=es$. ∎

Lemma 2.28**.**

If $R$ is a generalized weakly Rickart $*$-ring and $S$ is self adjoint subset of $R$, then $S^{\prime}$ is a weakly generalized Rickart $*$-ring.

Proof.

Let $x\in S^{\prime}$ and $GRP(x)=e$ in $R$. By Proposition 2.27, $se=es$ for all $s\in S$. Hence $e\in S^{\prime}$. Therefore $GRP(x)=e$ in $S^{\prime}$. Thus $S^{\prime}$ is a generalized weakly Rickart $*$-ring. ∎

3. Unitification of generalized weakly Rickart $*$-ring

Recall the definition of unitification of a $*$-ring given by Berberian [3]. Let $R$ be a $*$-ring. We say that $R_{1}$ is a unitification of $R$, if there exists a ring $K$, such that,

1. $K$ is an integral domain with involution (necessarily proper), that is, $K$ is a commutative $*$-ring with unity and without divisors of zero (the identity involution is permitted),

2. $R$ is a $*$-algebra over $K$ (i.e., $R$ is a left $K$-module such that, identically $1a=a,~\lambda(ab)=(\lambda a)b=a(\lambda b),~and~(\lambda a)^{*}=\lambda^{*}a^{*}$, for $\lambda\in K$ and $a,b\in R$).

3. $R$ is torsion free $K$-module (that is $\lambda a=0$ implies $\lambda=0$ or $a=0$).

Define $R_{1}=R\oplus K$ (the additive group direct sum), thus $(a,\lambda)=(b,\mu)$ means, by the definition that $a=b$ and $\lambda=\mu$, and addition in $R_{1}$, is defined by the formula $(a,\lambda)+(b,\mu)=(a+b,\lambda+\mu)$. Define $(a,\lambda)(b,\mu)=(ab+\mu a+\lambda b,\lambda\mu)$, $\mu(a,\lambda)=(\mu a,\mu\lambda)$, $(a,\lambda)^{*}=(a^{*},\lambda^{*})$. Evidently, $R_{1}$ is also a $*$-algebra over $K$, has unity element $(0,1)$ and $R$ is a $*$-ideal in $R_{1}$.

Berberian has given a partial solution to Problem 1 as follows.

Theorem 3.1** ([3, Theorem 1, page 31]).**

Let $R$ be a weakly Rickart $*$-ring. If there exists an involutory integral domain $K$ such that $R$ is a $*$-algebra over $K$ and it is a torsion-free $K$-module, then $R$ can be embedded in a Rickart $*$-ring with preservation of RP’s.

After 1972, there was little progress made toward the solution of Problem 1. In [14], Thakare and Waphare provided partial solutions, where the condition on the underlying weakly Rickart $*$-rings was relaxed in two distinct ways. For the solution of this open problem, Berberian used the condition that $R$ is a torsion-free left K-module, where K is an integral domain. Thakare and Waphare offered another solution in which the torsion-free condition was replaced with a different condition. They established the following.

Theorem 3.2** ([12, Theorem 2]).**

A weakly Rickart $*$-ring $R$ can be embedded into a Rickart $*$-ring, provided there exists a ring $K$ such that

1. (1)

$K$* is an integral domain with involution,* 2. (2)

$R$* is $*$-algebra over $K$, and* 3. (3)

For any $\lambda\in K-\{0\}$, there exist a projection $e_{\lambda}$ that is an upper bound for the set of left projections of the right annihilators of $\lambda$, that is if $x\in R$ and $\lambda x=0$ then $LP(x)\leq e_{\lambda}$.

Based on the theory developed in Section 2, we pose the following problem for generalized Rickart $*$-rings, similar to Problem 1.

Problem 2: Can every generalized weakly Rickart $*$-ring be embedded in a generalized Rickart $*$-ring? with preservation of $GRP$.

For a partial solution to Problem 2, the following results are required.

Proposition 3.3**.**

If $(a,0)\in R_{1}=R\oplus K$ then $(a,0)^{n}=(a^{n},0)$ for all $n\in\mathbb{N}$.

Proof.

We prove the result by using mathematical induction on $n$. Clearly result holds for $n=1$. Suppose result is true for $n=k$. That is $(a,0)^{k}=(a^{k},0)$. Consider $(a,0)^{k+1}=(a,0)(a,0)^{k}=(a,0)(a^{k},0)=(aa^{k},0)=(a^{k+1},0)$. Thus by method of induction $(a,0)^{n}=(a^{n},0)$ for all $n\in\mathbb{N}$. ∎

Proposition 3.4**.**

If $(a,\lambda)\in R_{1}$ and $n\in\mathbb{N}$ then $(a,\lambda)^{n}=(a^{n}+\binom{n}{1}a^{n-1}\lambda+\binom{n}{2}a^{n-2}\lambda^{2}+\cdots+\binom{n}{n-1}a\lambda^{n-1},\lambda^{n})$.

Proof.

We prove the result by using mathematical induction on $n$. Clearly result is true for $n=1$. Suppose the result is true for $n=k$. That is $(a,\lambda)^{k}=(a^{k}+\binom{k}{1}a^{k-1}\lambda+\cdots+\binom{k}{k-1}a\lambda^{k-1},\lambda^{k})$. Consider $(a,\lambda)^{k+1}=(a,\lambda)^{k}(a,\lambda)=(a^{k}+\binom{k}{1}a^{k-1}\lambda+\cdots+\binom{k}{k-1}a\lambda^{k-1},\lambda^{k})(a,\lambda)$ $=(a^{k+1}+\binom{k}{1}a^{k}\lambda+\cdots+\binom{k}{k-1}a^{2}\lambda^{k-1}+a\lambda^{k}+a^{k}\lambda+\binom{k}{1}a^{k-1}\lambda^{2}+\cdots+\binom{k}{k-1}a\lambda^{k},\lambda^{k+1})=(a^{k+1}+[\binom{k}{0}+\binom{k}{1}]a^{k}\lambda+\cdots+[\binom{k}{k-1}+\binom{k}{k}]a\lambda^{k},\lambda^{k+1})=(a^{k+1}+\binom{k+1}{1}a^{k}\lambda+\binom{k+1}{2}a^{k-1}\lambda^{2}+\cdots+\binom{k+1}{k}a\lambda^{k},\lambda^{k+1})$. Hence by induction, $(a,\lambda)^{n}=(a^{n}+\binom{n}{1}a^{n-1}\lambda+\binom{n}{2}a^{n-2}\lambda^{2}+\cdots+\binom{n}{n-1}a\lambda^{n-1},\lambda^{n})$ for all $n\in\mathbb{N}$. ∎

Lemma 3.5**.**

If a ring $R$ has weakly proper involution then the involution on $R_{1}$ is weakly proper.

Proof.

Since the involution in $R$ is weakly proper. Therefore for $x\in R$, $xx^{*}=0$ implies $x^{n}=0$ for some $n\in\mathbb{N}$. Let $(a,\lambda)\in R_{1}$ and $(a,\lambda)(a,\lambda)^{*}=0$. This gives $(a,\lambda)(a^{*},\lambda^{*})=0$. Therefore $(aa^{*}+\lambda^{*}a+\lambda a^{*},\lambda\lambda^{*})=0$. Since $K$ is an integral domain, $\lambda\lambda^{*}=0$ implies $\lambda=0$. Hence $(aa^{*},0)=0$. Therefore $aa^{*}=0$. Thus $a^{n}=0$ for some $n\in\mathbb{N}$. Hence $(a,\lambda)^{n}=(a,0)^{n}=(a^{n},0)=(0,0)=0$. So $R_{1}$ has weakly proper involution. ∎

In the following result, we provide a partial solution to Problem 2.

Theorem 3.6**.**

A generalized weakly Rickart $*$-ring can be embedded in a generalized Rickart $*$-ring provided there exists a ring $K$ such that

1. (1)

$K$* is an integral domain with involution.* 2. (2)

$R$* is $*$-algebra over $K$.* 3. (3)

For any nonzero $\lambda\in K$, there exists a projection $e_{\lambda}$ such that $\lambda x=0$ implies $GRP(x)\leq e_{\lambda}$.

Proof.

Let $R_{1}=R\oplus K$ (the additive group direct sum) with operations as defined above. First we prove that for any self-adjoint element $a\in R$ and $0\neq\lambda\in K$ there exists largest projection $g$ such that $(ag+\lambda g)^{m}=0$ for some $m\in\mathbb{N}$. Let $GRP(a)=e_{0}$. Then $a^{m}e_{0}=a^{m}$ and $a^{m}y=0$ implies $e_{0}y=0$ for some $m\in\mathbb{N}$. Let $e_{\lambda}$ be a projection which exists by the assumption (3). Let $e$ be the largest projection in $\{e_{0},e_{\lambda}\}$. Let $GRP(ae+\lambda e)=h$. Hence there exists $m\in\mathbb{N}$ such that $(ae+\lambda e)^{m}h=(ae+\lambda e)^{m}$ and $(ae+\lambda e)^{m}y=0$ implies $hy=0$. Now $e_{0}\leq e$ implies $e_{0}=e_{0}e=ee_{0}$. Let $g=e-h$. Since $(ae+\lambda e)^{m}e=(ae+\lambda e)^{m}$, we have $h\leq e$, that is $h=he=eh$. Therefore $eg=e-eh=e-h=g$. This gives $g\leq e$ and hence $g=eg=ge$. Thus $(ag+\lambda g)^{m}=(aeg+\lambda eg)^{m}=(ae+\lambda e)^{m}g=(ae+\lambda e)^{m}(e-h)=(ae+\lambda e)^{m}-(ae+\lambda e)^{m}h=0$. To prove $g$ is largest. Suppose $(ak+\lambda k)^{m}=0$. We have $e_{0}=e_{0}e=ee_{0}$. Since $a^{m}e_{0}=a^{m}$, $a^{m}e_{0}e=a^{m}e$, which implies $a^{m}e_{0}=a^{m}e$. Therefore $ea^{m}=a^{m}$, which gives $kea^{m}k=ka^{m}k$. Hence $(ke-k)a^{m}k=0$. Since $(ak+\lambda k)^{m}=0$, we have $(ke-k)\{-\binom{m}{1}a^{m-1}k\lambda k-\cdots-\lambda^{m}k\}=0$. Equating coefficient of $a^{m-1}k$, we get $\lambda m(ke-k)=0$. Therefore $\lambda(ke-k)=0$. Let $GRP(ke-k)=f$. Then $(ke-k)^{n}f=(ke-k)^{n}$ and $(ke-k)^{n}y=0$ implies $fy=0$. Since $(ke-k)^{n}e=0$, we have $fe=0$. Further $\lambda(ke-k)=0$ implies $\lambda(ke-k)^{n}=0$. Which gives $\lambda(ke-k)^{n}f=0$. Hence $(ke-k)^{n}\lambda f=0$ . Therefore $f(\lambda f)=0$ . Thus $\lambda f=0$. By (3) $GRP(f)\leq e_{\lambda}\leq e$. Therefore $f\leq e$, that is $f=fe=ef$ (since $GRP(f)=f$). Hence $f=0$. As $(ke-k)^{n}=(ke-k)^{n}f$ implies $(ke-k)^{n}=0$. But $(ke-k)^{n}=\pm(ke-k)$. Hence $\pm(ke-k)=0$ gives $ke=k$. So $(ae+\lambda e)^{m}k=(aek+\lambda ek)^{m}=(ak+\lambda k)^{m}=0$. Therefore $hk=0$. Hence $kg=k(e-h)=ke-kh=k-0=k$. Thus $k\leq g$. Hence $g$ is the largest projection such that $(ag+\lambda g)^{m}=0$. Since $(0,1)$ is the unity element of $R_{1}$. By above Proposition 2.20 it is enough to show that $R_{1}$ is a generalized weakly Rickart $*$-ring. Let $(a,\lambda)\in R_{1}$.

Suppose $\lambda=0$. Since $(a,0)\in R_{1}$, we have $a\in R$. As $R$ is a generalized weakly Rickart $*$-ring, $GRP(a)=e$ exists in $R$. That is for some $n\in\mathbb{N}$, $a^{n}e=a^{n}$; and for $y\in R$, $a^{n}y=0$ implies $ey=0$. We will prove that $GRP(a,0)=(e,0)$. Since $(a,0)^{n}(e,0)=(a^{n},0)(e,0)=(a^{n}e,0)=(a^{n},0)=(a,0)^{n}$. Suppose $(a,0)^{n}(b,\mu)=0$. Then $(a^{n},0)(b,\mu)=0$. This implies $(a^{n}b+\mu a^{n},0)=0$, that is $a^{n}b+\mu a^{n}=0$. This gives $a^{n}b+\mu a^{n}e=0$, and hence $a^{n}(b+\mu e)=0$. Since $a^{n}y=0$ implies $ey=0$, we have $e(b+\mu e)=0$. Therefore $eb+\mu e=0$. That is $(e,0)(b,\mu)=0$. Hence $GRP(a,0)=(e,0)$.

Now, suppose $\lambda\neq 0$. Then there exists a largest projection $g$ in $R$ such that $(ag+\lambda g)^{t}=0$ for some $t\in\mathbb{N}$. Note that $(-g,1)$ is a projection in $R_{1}$. We prove that $GRP(a,\lambda)=(-g,1)$. Consider $(a,\lambda)^{t}(-g,1)=(a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})(-g,1)=(-a^{t}g-\binom{t}{1}a^{t-1}\lambda g-\cdots-\binom{t}{t-1}a\lambda^{n-1}g-\lambda^{t}g+a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})=(-(ag+\lambda g)^{t}+a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})=(a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})=(a,\lambda)^{t}$. To prove $(a,\lambda)^{t}(b,\mu)=0$ implies $(-g,1)(b,\mu)=0$. Let $(a,\lambda)^{t}(b,\mu)=0$. Then $(a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})(b,\mu)=0$. This implies $\lambda^{t}\mu=0\Rightarrow\mu=0$ (since $\lambda\neq 0$). Therefore $(a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})(b,0)=0$.

To prove $(a,\lambda)^{t}(b,0)^{m}=0$ implies $(-g,1)(b,0)^{m}=0$ for some $m\in\mathbb{N}$. Let $GLP(b)=f$, then there exist $m\in\mathbb{N}$ such that $fb^{m}=b^{m}$; and for $y\in R$, $yb^{m}=0$ implies $yf=0$. Therefore $(a,\lambda)^{t}(b,0)^{m}=0$. This implies that $(a^{t}+\binom{t}{1}a^{t-1}\lambda+\cdots+\binom{t}{t-1}a\lambda^{t-1},\lambda^{t})(b^{m},0)=0$. Therefore $(\{a^{t}+\binom{t}{1}\lambda a^{t-1}+\cdots+\binom{t}{t-1}\lambda^{t-1}a\}b^{m}+\lambda^{t}b^{m},0)=0$. Hence $(\{a^{t}f+\binom{t}{1}\lambda a^{t-1}f+\cdots+\binom{t}{t-1}\lambda^{t-1}af+\lambda^{t}f\}b^{m},0)=0$. This gives $(af+\lambda f)^{t}b^{m}=0$. Since $yb^{m}=0$ implies $yf=0$, we have $(af+\lambda f)^{t}f=0$. This implies $(af+\lambda f)^{t}=0$. But $g$ is the largest projection such that $(ag+\lambda g)^{t}=0$. Therefore $f\leq g$, which gives $f(1-g)=0$. Hence $(-g,1)(b,0)^{m}=(-g,1)(b^{m},0)=(-gb^{m}+b^{m},0)=((1-g)b^{m},0)=((1-g)fb^{m},0)=(0,0)=0$.

Hence $GRP(a,\lambda)=(-g,1)$. Thus $R_{1}$ is a generalized Rickart $*$-ring. ∎

4. Parallelogram law, Generalized comparability and partial comparability

In this section, we prove that generalized Rickart $*$-ring satisfies parallelogram law. A sufficient condition is provided for a generalized Rickart $*$-ring to exhibit partial comparability. It is shown that a pair of projections in a generalized Rickart $*$-ring that satisfy the parallelogram law possesses orthogonal decomposition.

Proposition 4.1**.**

Let $R$ be a generalized Rickart $*$-ring such that $GLP(x)\sim GRP(x)$ for all $x\in R$. Then $R$ satisfies the parallelogram law.

Proof.

Let $x=e-ef=e(1-f)$. Then $e\vee f=f+GRP(e(1-f))$. Therefore $GRP(e(1-f))=e\vee f-f$. Also $GLP(e(1-f))=e-e\wedge f$. Since $GLP(x)\sim GRP(x)$, we have $GLP(e(1-f))\sim GRP(e(1-f))$. Therefore $e-e\wedge f\sim e\vee f-f$. Thus $R$ satisfies the parallelogram law. ∎

Projections $e$ and $f$ are said to be in a position $p^{\prime}$ in case $e\wedge(1-f)=0$ and $e\vee(1-f)=1$, that is $e$ and $1-f$ are complementary.

Proposition 4.2**.**

Let $R$ be a generalized Rickart $*$-ring and $e,~f$ are projections in $R$. Then the following are equivalent.

1. (1)

$e,f$* are in position $p^{\prime}$.* 2. (2)

$GLP(ef)=e$* and $GRP(ef)=f$.*

Proof.

Suppose $e,f$ are in position $p^{\prime}$. Therefore $e\wedge(1-f)=0$ and $e\vee(1-f)=1$. Note that $ef=e(1-(1-f))$. Since $e\vee f=f+GRP(e(1-f))$, we have $GRP(ef)=GRP(e(1-(1-f)))=e\vee(1-f)-(1-f)=e\vee(1-f)-1+f=f$. Similarly, $GL(ef)=GLP(e(1-(1-f)))=e-e\wedge(1-f)=e$. Conversely suppose $GLP(ef)=e$ and $GRP(ef)=f$. Therefore $GRP(e(1-(1-f)))=f$ implies $e\vee(1-f)-(1-f)=f$. This gives $e\vee(1-f)-1+f=f$ that is $e\vee(1-f)=1$. Similarly $GLP(ef)=e$ implies $e\wedge(1-f)=0$. Thus $e,f$ are in a position $p^{\prime}$. ∎

Proposition 4.3**.**

Let $R$ be a generalized Rickart $*$-ring. Then the following are equivalent.

1. (1)

$R$* satisfies the parallelogram law.* 2. (2)

If $e,f$ are in position $p^{\prime}$ then $e\sim f$.

Proof.

Suppose $R$ satisfies the parallelogram law. Let $e,f$ be projections in a position $p^{\prime}$. Thus $e\wedge(1-f)=0$ and $e\vee(1-f)=1$. By the parallelogram law $e-e\wedge f\sim e\vee f-f$. Replacing $f$ by $1-f$, we get $e-e\wedge(1-f)\sim e\vee(1-f)-(1-f)$. Hence $e-0\sim 1-1+f$, that is $e\sim f$. Conversely, suppose $e,f$ are in a position $p^{\prime}$ implies $e\sim f$. Let $e,f$ be a pair of projections. Let $GRP(ef)=f^{\prime}$ and $GLP(ef)=e^{\prime}$. Since $eff=ef$, we have $f^{\prime}\leq f$ that is $f^{\prime}=f^{\prime}f$. Similarly, since $eef=ef$, we have $e^{\prime}\leq e$ that is $e^{\prime}=e^{\prime}e$. Therefore $ef=e^{\prime}(ef)f^{\prime}=(e^{\prime}e)ff^{\prime}=e^{\prime}f^{\prime}$. Thus $GRP(e^{\prime}f^{\prime})=f^{\prime}$ and $GLP(e^{\prime}f^{\prime})=e^{\prime}$. By Proposition 4.2, $e^{\prime}$ and $f^{\prime}$ are in a position $p^{\prime}$. Therefore $e^{\prime}$ $\sim$ $f^{\prime}$. Note that $e\wedge f=e-GLP(e(1-f))$. Replacing $f$ by $1-f$, we get $e\wedge(1-f)=e-GLP(ef)=e-e^{\prime}$. Hence $e^{\prime}=e-e\wedge(1-f)$. Similarly $f^{\prime}=e\vee(1-f)-1+f$. Therefore $e^{\prime}\sim f^{\prime}$, which gives $e-e\wedge(1-f)\sim e\vee(1-f)-1+f=f-(1-e)\wedge f$. Hence $e-e\wedge(1-f)\sim f-(1-e)\wedge f$. Thus $R$ satisfies the parallelogram law. ∎

Projections $e$ and $f$ in a $*$-ring $R$ are said to be generalized comparable if there exists central projection $h$ such that $he\lesssim hf$ and $(1-h)f\lesssim(1-h)e$. We say that $R$ has generalized comparability (GC) if every pair of projections is generalized comparable. Projections $e$ and $f$ in a $*$-ring $R$ are said to be very orthogonal if there exists central projection $h$ such that $he=e$ and $hf=0$

Proposition 4.4**.**

If projection $e$ and $f$ are very orthogonal in a generalized Rickart $*$-ring $R$, then $e,f$ are orthogonal, $GRP(e)GLP(f)=0$ and $eRf=0$.

Proof.

Suppose $e$ and $f$ are very orthogonal. Therefore their exists central projection $h$ such that $he=e$ and $hf=0$. Hence $ef=hef=ehf=0$. Therefore $e$ and $f$ are orthogonal. Further, $GRP(e)=e$ and $GLP(f)=f$. Hence $GRP(e)GLP(f)=ef=0$. Also, $eRf=heRf=eRhf=0$. ∎

Example 4.5**.**

Orthogonal projections need not be very orthogonal. In $\mathbb{Z}_{12}$, the projections $0,1,4,$ and $9$ are all central. Since $2\cdot 6=0$, we have $2$ and $6$ are orthogonal. But $h\cdot 2=2$ and $h\cdot 6=0$ does not hold for any central projection $h$ in $\mathbb{Z}_{12}$. Therefore $2,6$ are not very orthogonal.

Projections $e$ and $f$ in a $*$-ring $R$ are said to be partially comparable if there exists nonzero projections $e_{0}$ and $f_{0}$ such that $e_{0}\leq e,f_{0}\leq f$ and $e_{0}\sim f_{0}$. We say that $R$ has partial comparability (PC) if $eRf\neq 0$ implies $e,f$ are partially comparable. A $*$-ring is said to have orthogonal GC if every pair of orthogonal projections is generalized comparable.

Proposition 4.6**.**

If $R$ is a generalized Rickart $*$-ring with $GRP(x)\sim GLP(x)$ for all $x\in R$ then $R$ has $PC$.

Proof.

Let $e$ and $f$ be projections in $R$ such that $eRf\neq 0$. Let $x=eaf\in eRf$ be such that $x\neq 0$. Let $GRP(x)=f_{0}$ and $GLP(x)=e_{0}$. Then there exists $n\in$ $\mathbb{N}$ such that $x^{n}f_{0}=x^{n}$; and for $y\in R$, $x^{n}y=0$ implies $f_{0}y=0$. Now $xf=x$ gives $x^{n}f=x^{n}$. Since $GRP(x)=f_{0}$, we have $f_{0}\leq f$. Similarly $e_{0}\leq e$. As $GLP(x)\sim GRP(x)$, we have $e_{0}\sim f_{0}$. Therefore there exist $e_{0},f_{0}$ such that $e_{0}\leq e$, $f_{0}\leq f$ and $e_{0}\sim f_{0}$. Therefore $R$ has $PC$. ∎

Example 4.7**.**

Let $R=M_{2}(\mathbb{Z}_{3})$ and $e=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$, $f=\begin{bmatrix}2&2\\ 2&2\end{bmatrix}\in R$. Let $A=\begin{bmatrix}a&b\\ c&d\end{bmatrix}\in R$ be such that $A^{*}A=e$ and $AA^{*}=f$. Therefore $\begin{bmatrix}a&c\\ b&d\end{bmatrix}$ $\begin{bmatrix}a&b\\ c&d\end{bmatrix}=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$ and $\begin{bmatrix}a&b\\ c&d\end{bmatrix}\begin{bmatrix}a&c\\ b&d\end{bmatrix}=\begin{bmatrix}2&2\\ 2&2\end{bmatrix}$. This gives $a^{2}+c^{2}=1,ab+cd=0,b^{2}+d^{2}=0$ and $a^{2}+b^{2}=2,ac+bd=2,c^{2}+d^{2}=2$. If $a=0$ then $c=1$ or $2$. In both cases $~b=0,~d=0$. If $a=1$ then $b=c=d=0$. If $a=2$ then $b=0,~c=0,~d=0$. Observe that none of the solution satisfy $a^{2}$+$b^{2}=2$. Hence $A^{*}A=e$ and $AA^{*}=f$ do not hold for any $A\in R$. Therefore $e\nsim f$.

Example 4.8**.**

Let $R=M_{2}$($\mathbb{Z}_{3}$), $e=\begin{bmatrix}1&0\\ 0&0\end{bmatrix}$, $f=\begin{bmatrix}2&2\\ 2&2\end{bmatrix}$, $1-e=\begin{bmatrix}0&0\\ 0&1\end{bmatrix}$, $1-f=\begin{bmatrix}2&1\\ 1&2\end{bmatrix}\in R$. Note that $e(1-f)=\begin{bmatrix}2&1\\ 0&0\end{bmatrix}$ , $ef=\begin{bmatrix}2&2\\ 0&0\end{bmatrix}$ , $f(1-e)=\begin{bmatrix}0&2\\ 0&2\end{bmatrix}$ , $e(1-e)=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$, $f(1-f)=\begin{bmatrix}0&0\\ 0&0\end{bmatrix}$. Hence $e,f,1-e,1-f$ are incomparable. For the pair $e,1-f$, we have $e-e\wedge(1-f)=e$ and $e\vee(1-f)-1+f=f$. But $e\nsim f$. Hence for $e,1-f$ parallelogram law does not hold. Note that $0,1$ are only central projections in $R$. Let $e,f$ as above. For $h=1,he\lnsim hf$ that is $e\lnsim f$, because if $e\lesssim f$ then $e\sim g\leq f$ this implies $g=0$ or $f=0$ but $e\sim 0$ gives $e=0$, a contradiction and $g=f$ gives $e\sim f$ a contradiction. Therefore $e\lnsim f$. For $h=0,(1-h)f\lnsim(1-h)e$ that is $f\lnsim e$ as above. Thus $R$ does not have $GC$. We have $eRf\neq 0$ but $e,f$ do not have non-zero sub-projections $e_{0},f_{0}$ such that $e_{0}\sim f_{0}$. In fact $e$ and $f$ are only non-zero sub-projections of $e,f$ respectively but $e\nsim f$. Thus $R$ does not have $PC$.

Proposition 4.9**.**

Let $R$ be a generalized Rickart $*$-ring satisfying parallelogram law. If $e,f$ are projections in $R$, then there exists orthogonal decomposition $e=e^{\prime}+e^{\prime\prime}$, $f=f^{\prime}+f^{\prime\prime}$ with $e^{\prime}\sim f^{\prime}$ and $ef^{\prime\prime}=fe^{\prime\prime}=0$.

Proof.

Let $GLP(ef)=e^{\prime}$ and $GRP(ef)=f^{\prime}$. Then $ef=e^{\prime}f^{\prime}$. Therefore $GLP(e^{\prime}f^{\prime})=e^{\prime}$ and $GRP(e^{\prime}f^{\prime})=f^{\prime}$. By Proposition 4.2, we have $e^{\prime}$ and $f^{\prime}$ are in position $p^{\prime}$. Therefore $e^{\prime}\sim f^{\prime}$. Let $e^{\prime\prime}=e-e^{\prime}$, $f^{\prime\prime}=f-f^{\prime}$. Then $e^{\prime\prime}(ef)=(e-e^{\prime})ef=ef-e^{\prime}ef=ef-ef=0$. Therefore $(e^{\prime\prime}e)f=0$. Since $e^{\prime\prime}\leq e$, we have $e^{\prime\prime}f=0$. Similarly $(ef)f^{\prime\prime}=0$ gives $ef^{\prime\prime}=0$. ∎

Disclosure statement: The authors report there are no competing interests to declare.

Acknowledgment: The authors are thankful to the anonymous referees for helpful comments and suggestions.