\documentstyle{SibMatJ} %\TestXML \topmatter \Author Khairnar \Initial A. \Sign Anil Khairnar \Email ask.agc\@mespune.in \Email anil\_maths2004\@yahoo.com \AffilRef 1 \Corresponding \endAuthor \Author More \Initial S. \Sign Sanjay More \Email sanjaymore71\@gmail.com \AffilRef 2 \endAuthor \Affil 1 \Division Department of Mathematics \Organization MES Abasaheb Garware College \City Pune \Country India \endAffil \Affil 2 \Division Department of Mathematics \Organization Prof. Ramkrishna More College \City Pune \Country India \endAffil \datesubmitted January 29, 2025\enddatesubmitted \daterevised July 19, 2025\enddaterevised \dateaccepted August 4, 2025\enddateaccepted \UDclass ??? %\?Primary 16W10; Secondary 16L45 \endUDclass \title On Generalized Rickart $*$-Rings \endtitle \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 %\?есть без артикля 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 %\?weakly generalized еще есть, много Rickart $*$-ring and provide a characterization of generalized Rickart $*$-rings in terms of weakly generalized Rickart $*$-rings. %\? иногда так пишут, иногда просто weakly Rickart 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 %\?pairs of projections in a~generalized Rickart $*$-ring possess %\? orthogonal decomposition. \endabstract \keywords generalized Rickart $*$-ring, generalized right projection, projections, generalized weakly Rickart $*$-ring \endkeywords \endtopmatter \head 1. Introduction \endhead Kaplansky [1] 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 [2--7]. A ring $R$ is said to be {\it reduced\/} if it does not contains %\?contain nonzero nilpotent element. %\? A ring $R$ is said to be {\it abelian\/} if its every idempotent element is central. Let $S$ be a nonempty subset of $R$. We write $r(S)= \{a \in R : s a = 0 \text{ for all } s \in S \}$, and is called %\? the {\it right annihilator\/} of $S$ in $R$, and $l(S)= \{a \in R \mid a s = 0 \text{ for all } s \in S \}$ is the { \it 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 {\it $*$-ring $R$\/} is a ring equipped with an involution $x \rightarrow x^* $, that is, an additive antiautomorphism of the period at most two. An element $e$ of a $*$-ring $R$ is called a~{\it 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 let $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 {\it 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 %\?есть без артикля $e$ such that $xe=x$ and $xy=0$ if and only if $ey=0$, called the {\it 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 {\it 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 [8], Berberian posed the following open problem. \proclaim\nofrills{Problem 1}{\rm:} %\?надо : Can every weakly Rickart $*$-ring be embedded in a Rickart $*$-ring with preservation of~$RP$'s? %\?$RP$s? \endproclaim In [8] Berberian provided a partial solution to this problem. Subsequently, in [9], the authors offered another partial solution. In [6], a more general partial solution to \Par*{Problem 1} is presented. Let $R$ be a $*$-ring. The projections $e$ and $f$ in $R$ are said to be {\it equivalent\/} (written as $e \sim f$), if there exists $w\in R$ such that $w^*w=e$ and $ww^*=f$ (see [8]). By [8, Proposition 7, p.~5], the relation $\sim$ is an equivalence relation on the set of projections in $R$. A $*$-ring is said to satisfy the {\it parallelogram law\/} if $e-e\wedge f \sim e\vee f-f$ for every pair of projection %\?projections $e$ and $f$ whose $e\wedge f$ and $e\vee f$ exists. %\? Projections $e$ and $f$ in a $*$-ring $R$ are said to be {\it partially comparable\/} if there exist nonzero projections $e_{0}$ and $f_{0}$ such that $e_{0} \le e$, $f_{0} \le f$, and $e_{0} \sim f_{0}$. We say that a $*$-ring $R$ has {\it partial comparability\/} $(PC)$ %\?PC прямо if $eRf \neq 0$ implies $e$ and $f$ are partially comparable. Let $R$ be a $*$-ring and let $e$ and $f$ be projections in $R$. We say that $e$ is {\it 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 {\it generalized comparable\/} if there exists central projection %\?артикль еще есть без $h$ such that $he \lesssim hf$ and $(1-h)f \lesssim (1-h)e$. We say that $R$ has {\it generalized comparability\/} (GC) %\?$(GC)$ то прямо, то нет if every pair of projection %\?projections in $R$ is generalized comparable. We say that a $*$-ring $R$ has {\it orthogonal GC\/} if every pair of orthogonal projections are %\?is generalized comparable. Projections $e$ and $f$ in a $*$-ring $R$ are said to be {\it very orthogonal\/} if there exists central projection $h$ such that $he = e$ and $hf = 0$. In [10], 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 [11], 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 %\?finite-dimensional and infinite-dimensional Banach $*$-algebras that are generalized Rickart $*$-rings but not Rickart $*$-rings. In \Sec*{Section 2} of this paper, we introduce the 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 %\?projections 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 \Sec*{Section 3}, we pose a problem for generalized Rickart $*$-rings analogous to \Par*{Problem 1} for Rickart $*$-rings and provide a partial solution. In \Sec*{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. %\?PC Furthermore, it is shown that pairs of projections in a generalized Rickart $*$-ring that satisfying %\? the parallelogram law possess orthogonal decomposition. \head 2. Weakly Generalized Rickart $*$-Rings \endhead It is evident that every Rickart $*$-ring is a generalized Rickart $*$-ring. In this section, we first recall examples and results from [11], 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. \demo{Example 2.1 \rm [11, Example 2.8]}\Label{E2.1} \Item (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. %\?еще есть \Item (ii) Let $G$ be a finite abelian $p$-group. Then the group algebra $T = F_pG$ is a finite commutative local ring with unique maximal ideal $\operatorname{rad}(T)$ having index of nilpotency $|G|$. Let $*$ be the involution on the group ring $T$ defined by $(\sum a_gg)^* = \sum a_gg^{-1}$. Hence $T$ is a generalized Rickart $*$-ring but not Rickart $*$-ring. %\? \Item (iii) Let $S$ and $T$ be the rings in \Par{E2.1}{(i)} and \Par{E2.1}{(ii)} respectively. Let $n$ be a~positive integer and let~$p$ be a~prime. Then the $*$-ring $R = S \oplus T \oplus \Bbb Z_{p^n}$ is a generalized Rickart $*$-ring that is not Rickart $*$-ring. %\?a \Item (iv) Let $T=\{a_0+a_1i+a_2j+a_3k : a_i\in \Bbb Z_2 \text{ for } i=1,2,3,4\}$ be the Hamilton quaternions over~$\Bbb Z_2$. Then $T$ is a commutative ring such that every element of $T$ is either invertible or nilpotent. For $\alpha=a_0+a_1i+a_2j+a_3k$ with $a_0,a_1,a_2,a_3\in \Bbb Z_2$, define $\alpha^*=a_0-a_1i-a_2j-a_3k$. Then $*$ is an involution for $T$. Then $T$ is a generalized Rickart $*$-ring but not a Rickart $*$-ring. \enddemo \demo{Definition 2.2 \rm[12, 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 defined as follows: $$ \align A_n(R) &=RI_n+ \sum_{l=2}^{[\frac{n}{2}]}RV_n^{l-1}+\sum_{i=1}^{[\frac{n+1}{2}]}\sum_{j=[\frac{n}{2}]+i}^{n}RE_{ij}, \\ B_n(R) &=RI_n+ \sum_{l=3}^{[\frac{n}{2}]}RV_n^{l-2}+\sum_{i=1}^{[\frac{n+1}{2}]+1}\sum_{j=[\frac{n}{2}]+i-1}^{n}RE_{ij}, \\ U_n(R) &=RI_n+ \sum_{i=1}^{[\frac{n-1}{2}]}\sum_{j=[\frac{n}{2}]+1}^{n}RE_{ij}+\sum_{j=[\frac{n-1}{2}]+2}^{n}RE_{[\frac{n-1}{2}]+1,j}, \\ S_n(R) &=RI_n+ \sum_{i m\}$. Then $R$ is a ring with component-wise %\?слитно operations multiplication and addition. Define the involution $*$ on $R$ as, for $x=(x_1,x_2,\cdots )\in R$, $x^*=(\bar{x_1},\bar{x_2}, %\?\overline 2 раза \cdots )$. %\?\dots всюду 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 $m\in \Bbb{N}$ such that $x_k=0$ for all $k > m$. Let us find $n\in \Bbb{N}$ and projection %\? $e \in R$ such that $x^ne=x^n$ and $x^ny=0$ 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 nonzero components, $e$ also has finitely many nonzero components. So $e \in R$. Further, $e_i\in \{0,1\}$ and $e_i=\bar{e_i}$, %\?\overline thus $e$ is a~projection in $R$. For each component $(xe)_i=x_ie_i$. 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 $x_iy_i=0$ 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 $(ey)_i=e_iy_i$. 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 $n\in \Bbb{N}$, $r(x^n)\neq eR$ 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, $r(x)=\{(y_1,0,y_3,y_4, \cdots )\mid y_i\in \Bbb{C}\}$. 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 $y_i=e_iz_i$ for some $z=(z_1,z_2, \cdots )\in R$. As $y_2=0$ for any $y\in r(x)$. %\? So $(ez)_2=e_2z_2=0$. 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 nonzero components. Hence, $r(x^n)\neq eR$ for any projection $e\in R$ and for any $n\in \Bbb{N}$. Thus, $R$ is not a~generalized Rickart $*$-ring. \enddemo In the following result, we prove that the center of a generalized weakly Rickart $*$-ring is itself a~generalized weakly Rickart $*$-ring. \proclaim{Proposition 2.22} The center of a generalized weakly Rickart $*$-ring is generalized weakly Rickart $*$-ring. %\?a \endproclaim \demo{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. %\?соединить со следующим предложением Therefore $GRP(x)=e$ exists in $R$. That is there exists $n \in \Bbb N$ such that $x^ne=x^n$; and $x^ny=0$ implies $ey=0 $. Hence there exists $n \in \Bbb N$ such that $e(x^n)^* =(x^n)^* $; and $y^*(x^n)^*=0$ implies $y^*e=0$. Since $x \in c(R)$, we have $x^n(r-re)=x^nr-x^nre=x^nr-rx^ne=x^nr-rx^n =x^nr-x^nr=0$. 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. \qed\enddemo The involution $*$ of a ring $R$ is called {\it weakly proper\/} if for any $ x\in R $, $ xx^*=0$ implies $x^n=0$ for some $ n\in\Bbb N$. \proclaim{Proposition 2.23}\Label{P2.23} Let $R$ be a generalized weakly Rickart $*$-ring. Then, \Item (1) for each $ x\in R$ there exists $n \in \Bbb N$ such that $r(x^n) \cap(x^*)^nR=\{0\}$; \Item (2) the involution $*$ is weakly proper. \endproclaim \demo{Proof} \Par{P2.23}{(1)}: Let $x \in R$. Since $R$ is a generalized weakly Rickart $*$-ring, there exist $n\in\Bbb N$ and a projection $e \in R$ such that $x^ne=x^n$; and for $y \in R$, $x^ny=0$ implies $ey=0$. We prove that $r(x^n) \cap (x^*)^nR=\{0\}$. Let $y\in r(x^n)\cap (x^*)^nR$. Therefore $x^ny=0$ and $y=(x^*)^ns$ for some $ s\in R$. This gives $y=(x^*)^ns=(x^n)^*s=(x^ne)^*s=e(x^n)^*s=e(x^*)^ns=ey=0$. Hence $r(x^n)\cap (x^*)^nR=\{0\}$. \Par{P2.23}{(2)}: Let $ xx^*=0$. Therefore $x^n(x^*)^n=0$. This gives $e(x^*)^n=0$. Hence $e(x^n)^*=0$ implies $(x^ne)^*=0$, and this gives $(x^n)^*=0$. Thus $x^n=0$. \qed\enddemo \proclaim{Corollary 2.24 \rm [11, Proposition 2.11]} Let $R$ be a generalized Rickart $*$-ring. Then \Item (i) for each $x\in R$, there exists an integer $n \geq 1$ such that $r(x^n) \cap (x^*)^nR = 0 $; \Item (ii) the involution $*$ is weakly proper. \endproclaim The following result provides the characterization of a generalized weakly Rickart $*$-ring. \proclaim{Proposition 2.25} The following conditions on a $*$-ring $R$ are equivalent: \Item (1) $R$ is generalized weakly Rickart $*$-ring. %\?a \Item (2) The involution $*$ is weakly proper and for every $ x\in R$ there exist $n\in\Bbb N$ and a projection $e$ in $R$ such that $r(x^n)=r(e)$. \endproclaim \demo{Proof} Suppose $R$ is a generalized weakly Rickart $ *$-ring. By \Par*{Proposition 2.23}, involution %\?есть без артикля on $R$ is weakly proper. Let $ x\in R $ and $GRP(x)=e$. Let $y\in r(x^n)$. Therefore $x^ny=0$ 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 $x^ny=x^ney=x^n0=0$. Hence $y\in r(x^n)$. Therefore $r(e) \subseteq r(x^n)$. Thus $r(x^n)=r(e)$. Conversely, suppose that involution $*$ %\? is weakly proper and for every $ x\in R$ there exist $n\in\Bbb N$ 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 %\?that $e(1-e)=0$ implies $1-e\in r(x^n)$. Therefore $x^n(1-e)=0$ implies $x^n=x^ne$. If $x^ny=0$ 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. \qed\enddemo In the following result, we prove that the corner of a generalized weakly Rickart $*$-ring is itself a~generalized weakly Rickart $*$-ring. \proclaim{Proposition 2.26} Let $R$ be a $*$-ring and let $e$ be a projection in $R$. If $R$ is a generalized weakly Rickart $*$-ring then so is $eRe$. \endproclaim \demo{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 exists $n \in \Bbb N$ such that $x^nf=x^n$; and for $y\in R$, $x^ny=0$ implies $fy=0$. Since $x\in eRe$, we have $x=exe$. So $x^ne=(exe)^ne=(exe)^n=x^n$. Therefore $x^n(e-f) =x^ne-x^nf=x^ne-x^n$. 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 $x^nf=x^n$. 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. \qed\enddemo \proclaim{Proposition 2.27} Let R be a generalized weakly Rickart $*$-ring and let $S$ be a self-adjoint subset of $R$ and $x\in S'$. If $GRP(x)=e$ then $ se=ese=es $ for all $ s\in S $. \endproclaim \demo{Proof} Since $x\in S'$, we have $xs=sx$ for all $s\in S$. As $GRP(x)=e$. %\? Then there exists $ n\in\Bbb N$ such that $x^ne=x^n$; and for $y \in R$, $x^ny=0$ implies $ey=0$. We have $x^ns=sx^n$. Now $x^n(se-es)=x^nse-x^nes=sx^ne-x^ns=sx^n-x^ns=0$. Therefore $e(se-es)=0$ implies $ese=es$. Replacing $s$ by $s^*$ we get $es^*e=es^*$. Therefore $ese=se$. Thus $se=ese=es$. \qed\enddemo \proclaim{Lemma 2.28} If $R$ is a generalized weakly Rickart $ *$-ring and $S$ is self adjoint subset %\?a self-adjoint subset of $R$, then $ S' $ is a weakly generalized Rickart $ *$-ring. \endproclaim \demo{Proof} Let $ x\in S' $ and $GRP(x)=e $ in $R$. By \Par*{Proposition 2.27}, $se=es$ for all $ s\in S$. Hence $e\in S'$. Therefore $GRP(x)=e $ in $S'$. Thus $S'$ is a generalized weakly Rickart $*$-ring. \qed\enddemo \head 3. Unitification of a Generalized Weakly Rickart $*$-Ring %\?артикль a поставила \endhead Recall the definition of unitification of a $*$-ring given by Berberian [8]. Let $R$ be a $*$-ring. We say that $R_1$ is a unitification of $R$, if there exists a ring $K$ such that \Item (1) %1) $K$ is an integral domain with involution (necessarily proper), that is, $K$ is a commutative $*$-ring with unity and without divisors of zero %\?zero divisors (the identity involution is permitted). \Item (2) %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 $a,b \in R$). \Item (3) % 3) $R$ is torsion free $K$-module %\?a (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 \Par*{Problem 1} as follows. \proclaim{Theorem 3.1 \rm[8, Theorem 1, p.~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. %\? \endproclaim After 1972, there was little progress made toward the solution of \Par*{Problem 1}. In [9], 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. \proclaim{Theorem 3.2 \rm[9, Theorem 2]} A weakly Rickart $*$-ring $R$ can be embedded into a Rickart $*$-ring, provided there exists a ring $K$ such that \Item (1) $K$ is an integral domain with involution, \Item (2) $R$ is $*$-algebra %\?a over $K$, and %\? \Item (3) For %\?for any $\lambda \in K-\{0\}$, there exists 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 $LP(x)\leq e_\lambda$. \endproclaim Based on the theory developed in \Sec*{Section 2}, we pose the following problem for generalized Rickart $*$-rings, similar to \Par*{Problem 1}. \proclaim\nofrills{Problem 2}{\rm:} Can every generalized weakly Rickart $*$-ring be embedded in a generalized Rickart $*$-ring? %\? with preservation of $GRP$. %\? сюда ? \endproclaim For a partial solution to \Par*{Problem 2}, the following results are required. \proclaim{Proposition 3.3} If $(a,0)\in R_1 = R\oplus K$ then $(a,0)^n=(a^n,0)$ for all $n\in\Bbb N$. \endproclaim \demo{Proof} We prove the result by using mathematical induction %\?the on $n$. Clearly the result holds for $n=1$. Suppose the 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 $n\in\Bbb N$. \qed\enddemo \proclaim{Proposition 3.4} If $(a,\lambda)\in R_1$ and $n \in \Bbb N$ then $(a,\lambda)^n=(a^n+\binom{n}{1} a^{n-1}\lambda+\binom{n}{2} a^{n-2}\lambda^2+\dots+\binom{n}{n-1}a\lambda^{n-1}, \lambda^n)$. \endproclaim \demo{Proof} We prove the result by using mathematical induction %\?the on $n$. Clearly the result is true for $n=1$. Suppose the result is true for $n=k$. That is $(a,\lambda)^k=(a^k+\binom{k}{1}a^{k-1}\lambda+\dots+\binom{k}{k-1}a\lambda^{k-1},\lambda^k)$. Consider $(a,\lambda)^{k+1}=(a,\lambda)^k(a,\lambda) =(a^k+\binom{k}{1}a^{k-1}\lambda+\dots+ \binom{k}{k-1}a\lambda^{k-1},\lambda^k) (a,\lambda) $ $=(a^{k+1}+\binom{k}{1}a^k\lambda+\dots+\binom{k}{k-1}a^2\lambda^{k-1}+a\lambda^k +a^k\lambda+\binom{k}{1}a^{k-1}\lambda^2+\dots+\binom{k}{k-1}a\lambda^k,\lambda^{k+1}) =(a^{k+1}+[\binom{k}{0}+\binom{k}{1}]a^k\lambda+\dots+[\binom{k}{k-1}+\binom{k}{k}]a\lambda^k,\lambda^{k+1}) =(a^{k+1}+\binom{k+1}{1}a^k\lambda+\binom{k+1}{2}a^{k-1}\lambda^2+\dots+\binom{k+1}{k}a\lambda^k,\lambda^{k+1})$. Hence by induction, $(a,\lambda)^n=(a^n+\binom{n}{1}a^{n-1}\lambda+\binom{n}{2}a^{n-2}\lambda^2+\dots+\binom{n}{n-1}a\lambda^{n-1},\lambda^n)$ for all $n\in\Bbb N$. \qed\enddemo \proclaim{Lemma 3.5} If a ring $R$ has weakly proper involution then the involution on $R_1$ is weakly proper. \endproclaim \demo{Proof} Since the involution in $R$ is weakly proper. %\? Therefore for $x\in R$, $xx^*=0 $ implies $x^n=0$ for some $n\in\Bbb N$. Let $(a,\lambda) \in R_1$ and $(a,\lambda)(a,\lambda)^*=0$. This gives $(a,\lambda)(a^*,\lambda^*)=0$. So $(aa^*+\lambda^* a+\lambda a^*,\lambda\lambda^*)=0$. 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 $n\in\Bbb N$. Hence $(a,\lambda)^n=(a,0)^n=(a^n,0)=(0,0)=0$. So $R_1$ has weakly proper involution. \qed\enddemo In the following result, we provide a partial solution to \Par*{Problem 2}. \proclaim{Theorem 3.6}\Label{T3.6} A generalized weakly Rickart $*$-ring can be embedded in a generalized Rickart $*$-ring provided there exists a ring $K$ such that \Item (1) $K$ is an integral domain with involution. \Item (2) $R$ is $*$-algebra over $K$. %\?a \Item (3) For any nonzero $\lambda\in K$, there exists a projection $e_\lambda$ such that $\lambda x=0$ implies $GRP(x)\leq e_\lambda$. \endproclaim \demo{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 $0 \ne \lambda \in K$ there exists a largest projection $g$ such that $(ag+\lambda g)^m=0$ for some $m\in\Bbb N $. Let $GRP(a)=e_0$. Then $a^me_0=a^m$ and $a^my=0$ implies $e_0y=0$ for some $m\in\Bbb N$. Let $e_\lambda$ be a projection which %\?that exists by assumption \Par{T3.6}{(3)}. Let $e$ be the largest projection in $\{e_0,e_\lambda\}$. Let $GRP(ae+\lambda e)=h$. Hence there exists $m\in\Bbb N$ such that $(ae+\lambda e)^mh=(ae+\lambda e)^m$ and $(ae+\lambda e)^my=0$ implies $hy=0$. Now $e_0\leq e$ implies $e_0=e_0e=ee_0$. Let $g = e-h$. Since $(ae+\lambda e)^me=(ae+\lambda e)^m$, 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 $(ag+\lambda g)^m=(a eg+\lambda eg)^m=(ae+\lambda e)^mg =(ae+\lambda e)^m(e-h)=(ae+\lambda e)^m-(ae+\lambda e)^mh=0$. To prove that $g$ is largest. %\? Suppose $(ak+\lambda k)^m=0$. We have $e_0=e_0e=ee_0$. Since $a^me_0=a^m$, $a^me_0e=a^me$, which implies $a^me_0=a^me$. Therefore $ea^m=a^m$, which gives $kea^mk=ka^mk$. Hence $(ke-k)a^mk=0$. Since $(ak+\lambda k)^m=0$, we have $(ke-k)\{-\binom{m}{1}a^{m-1}k\lambda k-\dots-\lambda^mk\}=0$. 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)^nf= (ke-k)^n $ and $(ke-k)^ny=0$ implies $fy=0$. Since $(ke-k)^ne=0$, we have $fe=0$. Further $\lambda (ke-k)=0$ implies $\lambda (ke-k)^n=0$. Which gives %\? $\lambda (ke-k)^nf=0$. Hence $ (ke-k)^n\lambda f=0$. Therefore $f(\lambda f)=0$. Thus $\lambda f=0$. By \Par{T3.6}{(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 $(ke-k)^n= (ke-k)^nf$ 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 the above \Par*{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 $n\in \Bbb N$, $a^ne=a^n$; and for $y \in R$, $a^ny=0$ implies $ey=0$. We will prove that $GRP(a,0)=(e,0)$. Since $(a,0)^n(e,0)=(a^n,0)(e,0)=(a^ne,0)=(a^n,0)=(a,0)^n$. %\? Suppose $(a,0)^n(b,\mu)=0$. Then $(a^n,0)(b,\mu)=0$. This implies $(a^nb+\mu a^n,0)=0$, that is, $ a^nb+\mu a^n=0$. This gives $a^nb+\mu a^ne=0$, and hence $a^n (b+\mu e)=0$. Since $a^ny=0$ 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 $\lambda\ne 0$. Then there exists a largest projection $g$ in $R$ such that $(ag+\lambda g)^t=0$ for some $t\in\Bbb N$. Note that $(-g,1)$ is a projection in $R_1$. We prove that $GRP(a,\lambda)=(-g,1)$. Consider $(a,\lambda)^t(-g,1)=(a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda ^{t-1},\lambda^t)(-g,1) =(-a^tg-\binom {t}{1}a^{t-1}\lambda g-\dots -\binom {t}{t-1}a\lambda^{n-1}g-\lambda^tg+a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)=(-(ag+\lambda g)^t+a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)=(a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)=(a,\lambda)^t$. To prove $(a,\lambda)^t(b,\mu)=0$ implies $(-g,1)(b,\mu)=0$. %\? еще есть Let $(a,\lambda)^t(b,\mu)=0$. Then $(a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)(b,\mu)=0$. This implies $ \lambda^t\mu=0\Rightarrow\mu=0$ (since $\lambda\ne 0$). Therefore $(a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)(b,0)=0$. To prove $(a,\lambda)^t(b,0)^m=0$ implies $(-g,1)(b,0)^m=0$ for some $m \in \Bbb N$. Let $GLP(b)=f$. Then there exists $m \in \Bbb N$ 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 $(a^t+\binom {t}{1}a^{t-1}\lambda+\dots+\binom {t}{t-1}a\lambda^{t-1},\lambda^t)(b^m,0)=0$. Therefore $(\{a^t+\binom {t}{1}\lambda a^{t-1}+\dots+\binom {t}{t-1}\lambda^{t-1}a\}b^m+\lambda^tb^m,0)=0$. Hence $(\{a^tf+\binom {t}{1}\lambda a^{t-1}f+\dots+\binom {t}{t-1}\lambda^{t-1}af+\lambda^tf\}b^m,0)=0$. This gives $(af+\lambda f)^tb^m=0$. Since $yb^m=0$ implies $yf=0$, we have $(af+\lambda f)^tf=0$. 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. \qed\enddemo \head 4. Parallelogram Law, Generalized Comparability, and Partial Comparability \endhead In this section, we prove that generalized Rickart $*$-ring %\?a satisfies the 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. \proclaim{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. \endproclaim \demo{Proof} Let $x=e - ef =e(1-f)$. Then $e\vee f=f + GRP(e(1-f))$. Hence $ 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. \qed\enddemo Projections $e$ and $f$ are said to be in a %\? то с артиклем, то без {\it position $p'$\/} in case $e \wedge (1-f)=0$ and $e \vee (1-f)=1$, that is, $e$ and $1-f$ are complementary. \proclaim{Proposition 4.2} Let $R$ be a generalized Rickart $ *$-ring and let $e$ and $f$ be projections in $R$. Then the following are equivalent: \Item (1) $e$ and $f$ are in position $p'$. \Item (2) $GLP(ef)=e$ and $GRP(ef)=f$. \endproclaim \demo{Proof} Suppose $e$ and $f$ are in position $p'$. %\? 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$ %\?, и еще есть that is, $e \vee (1-f) = 1$. Similarly $GLP(ef) = e$ implies $e \wedge (1-f) = 0$. Thus $e$ and $f$ are in a position $p'$. \qed\enddemo \proclaim{Proposition 4.3} Let $R$ be a generalized Rickart $*$-ring. Then the following are equivalent: \Item (1) $R$ satisfies the parallelogram law. \Item (2) If $e$ and $f$ are in position $p'$ then $e \sim f$. \endproclaim \demo{Proof} Suppose $R$ satisfies the parallelogram law. Let $e$ and $f$ be projections in a position $p'$. 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 $e$ and $f$ are in a position $p'$ implies $e \sim f$. Let $e,f$ be a pair of projections. Let $GRP(ef) = f'$ and $GLP(ef) = e'$. Since $ef f = ef$, we have $f' \le f$ %\?, that is, $f'=f'f$. Similarly, since $e ef = ef$, we have $e' \le e$ %\?, that is, $e'=e'e$. Therefore $ef=e'(ef)f'=(e'e)ff'=e'f'$. Thus $ GRP(e'f') = f'$ and $GLP(e'f') = e'$. By \Par*{Proposition 4.2}, $e'$ and $f'$ are in a position $p'$. Therefore $e'$ $\sim$ $f'$. 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'$. Hence $e' = e-e\wedge(1-f)$. Similarly $f' = e\vee(1-f)-1+f$. Therefore $e' \sim f'$, 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. \qed\enddemo Projections $e$ and $f$ in a $*$-ring $R$ are said to be generalized comparable if there exists central projection $h$ %\?артикль such that $he\lesssim hf$ and $(1-h)f\lesssim (1-h)e$. We say that $R$ has generalized comparability (GC) %\?уже было сокращение 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 $h$ %\?артикль such that $he=e$ and $hf=0$. \proclaim{Proposition 4.4} If projections $e$ and $f$ are very orthogonal in a generalized Rickart $*$-ring $R$, then $e$ and $f$ are orthogonal, $GRP(e) GLP(f)=0$, and $eRf=0$. \endproclaim \demo{Proof} Suppose $e$ and $f$ are very orthogonal. Therefore there 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, $eRf = heRf = eRhf = 0$. \qed\enddemo \demo{Example 4.5} Orthogonal projections need not be very orthogonal. In $\Bbb{Z}_{12}$, the projections $0$, $1$, $4$, and $9$ are all central. Since $2\cdot6 = 0$, we have that $2$ and $6$ are orthogonal. But $h\cdot2 = 2$ and $h\cdot6 = 0$ does not hold for any central projection $h$ in $\Bbb{Z}_{12}$. Therefore, $2$ and $6$ are not very orthogonal. \enddemo Projections $e$ and $f$ in a $*$-ring $R$ are said to be partially comparable if there exist 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) %\?(PC) уже было сокращение, то прямо, то нет if $eRf\neq 0$ implies that $e$ and $f$ are partially comparable. A $*$-ring is said to have orthogonal GC if every pair of orthogonal projections is generalized comparable. \proclaim{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. \endproclaim \demo{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 $n \in \Bbb N$ 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 $f_{0} \le f$. Similarly $e_{0}\le e$. As $GLP(x) \sim GRP(x)$, we have $e_{0} \sim f_{0}$. Therefore there exist $e_{0}$ and $f_{0}$ such that $e_{0} \le e$, $f_{0} \le f$, and $e_{0} \sim f_{0}$. Therefore $R$ has $PC$. \qed\enddemo \demo{Example 4.7} Let $R=M_{2}(\Bbb{Z}_{3})$ and $e= \bmatrix 1 & 0\\ 0 & 0 \endbmatrix$, $f= \bmatrix 2 & 2\\ 2 & 2 \endbmatrix \in R$. Let $A=\bmatrix a & b\\ c & d \endbmatrix \in R$ be such that $A^*A=e$ and $AA^*=f$. Therefore $\bmatrix a & c\\ b & d \endbmatrix$ $\bmatrix a & b\\ c & d \endbmatrix= \bmatrix 1 & 0\\ 0 & 0 \endbmatrix$ and $\bmatrix a & b\\ c & d \endbmatrix \bmatrix a & c\\ b & d \endbmatrix =\bmatrix 2 & 2\\ 2 & 2 \endbmatrix$. 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 $b=0$ and $d=0$. If $a=1$ then $b=c=d=0$. If $a=2$ then $b=0$, $c=0$, and $d=0$. 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$. \enddemo \demo{Example 4.8} Let $R=M_{2}$($\Bbb{Z}_{3}$), $e=\bmatrix 1 & 0\\ 0 & 0 \endbmatrix$, $f=\bmatrix 2 & 2\\ 2 & 2 \endbmatrix$, $1-e=\bmatrix 0 & 0\\ 0 & 1 \endbmatrix$, $1-f=\bmatrix 2 & 1\\ 1 & 2 \endbmatrix \in R$. Note that $e(1-f)=\bmatrix 2 & 1\\ 0 & 0 \endbmatrix$, $ef=\bmatrix 2 & 2\\ 0 & 0 \endbmatrix$, $f(1-e)=\bmatrix 0 & 2\\ 0 & 2 \endbmatrix$, $e(1-e)=\bmatrix 0 & 0\\ 0 & 0 \endbmatrix$, $f(1-f)=\bmatrix 0 & 0\\ 0 & 0 \endbmatrix$. Hence $e$, $f$, $1-e$, and $1-f$ 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 $e, 1-f$ the parallelogram law does not hold. Note that $0$ and $1$ are only central projections in $R$. Let $e$ and $f$ be as above. For $h=1, he \lnsim hf$ that is, %\? $e \lnsim f$, because if $e \lesssim f $ then $e \sim g \le f$ 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 $e$ and $f$ do not have nonzero subprojections $e_{0}$ and $f_{0}$ such that $e_{0} \sim f_{0}$. In fact $e$ and $f$ are only nonzero subprojections of $e$ and $f$ respectively but $e \nsim f$. Thus $R$ does not have $PC$. \enddemo \proclaim{Proposition 4.9} Let $R$ be a generalized Rickart $*$-ring satisfying the parallelogram law. If $e$ and~$f$ are projections in $R$, then there exists orthogonal decomposition $e=e'+e''$, $f=f'+f''$ with $e' \sim f'$ and $ef''=fe''=0$. \endproclaim \demo{Proof} Let $GLP(ef)=e'$ and $GRP(ef)= f'$. Then $ef= e'f'$. Hence $ GLP(e'f')= e'$ and $GRP(e'f')= f'$. 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