\documentstyle{SibMatJ} % \TestXML \topmatter \Author Paul \Initial R. \Sign Rinki Paul \Email rinkipaul2k\@gmail.com \AffilRef 1 \endAuthor \Author Sarkar \Initial P. \Sign Paltu Sarkar \Email pal2\_math\_796\@nbu.ac.in \AffilRef 1 \Corresponding \endAuthor \Affil 1 \Division Department of Mathematics \Organization University of North Bengal \City West Bengal \Country India \endAffil \datesubmitted January 8, 2026\enddatesubmitted \dateaccepted January 16, 2026\enddateaccepted \UDclass ??? %\?16U80, 16W25, 47B47 \endUDclass \thanks The first author express %\? her thanks to the Department of Science and Technology (DST), Govt of India for complete financial support %\? through INSPIRE fellowship (no. DST/INSPIRE Fellowship/2019/IF190653, dated 19.08.2021). \endthanks \title An Investigation on Lie Ideals in $rp$-Algebra Using $d$-Derivation %\?артикли \endtitle \abstract In this investigation, $\Cal{A}$ represents an associative algebra over a ring $\Cal{R}$ with unity (not necessarily commutative), $\Cal{U}$ %\?is a Lie ideal of $\Cal{A}$, and $\Cal{D}_d$ %\?is a $d$-derivation on $\Cal{A}$. An additive mapping~$\Cal{D}_d$ from $\Cal{A}$ to itself is said to be a $d$-derivation, if there exists a ring derivation $d$ on $\Cal{R}$ such that $\Cal{D}_d(xy)=\Cal{D}_d(x)y+x\Cal{D}_d(y)$ and $\Cal{D}_d(rx)=d(r)x+r\Cal{D}_d(x)$ for all $r\in \Cal{R}$ and $x,y\in \Cal{A}$. An $rp$-algebra $\Cal{A}$ over a~ring~$\Cal{R}$ is an associative algebra which contains no divisors of zero and for any $r\in \Cal{R}$ and $p\in \Cal{A}$, $r\Cal{A}p=\{0\}$ implies either $r=0$ or $p=0$. By admitting $d$-derivation $\Cal{D}_d$ on $\Cal{A}$, we present the necessary and sufficient identities for commutativity of Lie ideals in $rp$-algebra. %\?есть без артикля We also examine the relationship among the center $\Cal{Z}(\Cal{A})$ and the centralizers $C_\Cal{A}(\Cal{D}_d(\Cal{U}))$ and $C_\Cal{A}(\Cal{U})$ in $rp$-algebra. Moreover, we explore the structures of Lie ideals in $rp$-algebra by restricting $d$-derivation %\?не везде артикли $\Cal{D}_d$ to $\Cal{U}$, which acts as a homomorphism (antihomomorphism, Jordan homomorphism) on $\Cal{U}$. We also address the commuting and co-commuting concepts in this Lie structure formation. In the end, %\?At the end, we conclude our investigation by discussing Posner and Herstein's results in $rp$-algebra related to Lie ideals. \endabstract \keywords algebra, $d$-derivation, $rp$-algebra, Lie ideal \endkeywords \endtopmatter \let\mathscr\goth \head 1. Introduction \endhead Let $\Cal{R}$ be a ring with unity (not necessarily commutative) and let $\Cal{A}$ be an associative algebra over a~ring $\Cal{R}$ with center $\Cal{Z}(\Cal{A})$. An $\Cal{R}$-submodule $\Cal{U}$ of an associative algebra $\Cal{A}$ over a ring $\Cal{R}$ is said to be a~Lie ideal (Lie subalgebra) %\?нет \it if $[\Cal{U},\Cal{A}]\subseteq \Cal{U}$ (respectively $[\Cal{U},\Cal{U}]\subseteq \Cal{U}$). A Lie ideal $\Cal{U}$ of $\Cal{A}$ is said to be a~square closed Lie ideal if $u^2\in \Cal{U}$ for all $u\in \Cal{U}$. A Lie ideal $\Cal{U}$ is said to be noncentral on $\Cal{A}$ if $\Cal{U}\nsubseteq \Cal{Z}(\Cal{A})$. An additive mapping $\mathscr{D}$ from a ring $\Cal{R}$ into itself is said to be a derivation if $\mathscr{D}(ab)=\mathscr{D}(a)b+a\mathscr{D}(b)$ for all $a,b\in \Cal{R}$. Bre{\v{s}}ar [1] introduced a broader class, known as generalized ring derivation, and later, Hvala [2] carried out an algebraic analysis of this notion in prime rings. A generalized derivation is an additive mapping $\mathscr{F}$ from $\Cal{R}$ into $\Cal{R}$ if there exists a derivation $\mathscr{D}$ on $\Cal{R}$ such that $\mathscr{F}(xy)=\mathscr{F}(x)y+x\mathscr{D}(y)$ for all $x,y\in \Cal{R}$. A multiplicative derivation of $\Cal{R}$ is a mapping $\mathscr{D}$ from $\Cal{R}$ into itself which is not necessarily additive but satisfies $\mathscr{D}(ab)=\mathscr{D}(a)b+a\mathscr{D}(b)$ for all $a,b\in \Cal{R}$. This multiplicative notion was further extended to a multiplicative (generalized)-derivation in such a way that the map %\?то map, то mapping по статье $\mathscr{F}$, not necessarily additive, satisfies $\mathscr{F}(xy)=\mathscr{F}(x)y+x\mathscr{D}(y)$ for all $x,y\in \Cal{R}$, where $\mathscr{D}$ is any map (need not be a derivation or even additive). For further references, see [3--7]. The symbols $[a,b]$ and $a\circ b$ stand for commutator and anticommutator %\? артикли respectively of elements $a,b\in \Cal{R}$. % Let $\Cal{S}$ be a nonempty subset of a ring $\Cal{R}$. A mapping $\mathscr{F}:\Cal{R}\rightarrow \Cal{R}$ is said to be commuting (centralizing) on $\Cal{S}$ if $[\mathscr{F}(x),x]=0$ (respectively $[\mathscr{F}(x),x]\in \Cal{Z}(\Cal{R}))$ for all $x\in \Cal{S}$. Two mappings $\mathscr{F}:\Cal{R}\rightarrow \Cal{R}$ and $\mathscr{G}:\Cal{R}\rightarrow \Cal{R}$ are said to be co-commmuting (co-centralizing) on $\Cal{S}$ if $\mathscr{F}(x)x-x\mathscr{G}(x)=0$ (respectively $\mathscr{F}(x)x-x\mathscr{G}(x)\in \Cal{Z}(\Cal{R}))$ for all $x\in \Cal{S}$. Moreover, an additive map $\mathscr{F}:\Cal{R}\rightarrow \Cal{R}$ acts as a homomorphism (antihomomorphism, Jordan homomorphism) on a subset $\Cal{S}$ of $\Cal{R}$ if $\mathscr{F}(xy)=\mathscr{F}(x)\mathscr{F}(y)$ (respectively $\mathscr{F}(xy)=\mathscr{F}(y)\mathscr{F}(x)$, $\mathscr{F}(x\circ y)=\mathscr{F}(x)\circ \mathscr{F}(y))$ for all $x,y\in \Cal{S}$. The study of commuting and centralizing maps on prime rings was initiated by Posner [8], and this line of work %\? has since been extended by many authors in diverse ways. In derivation theory, Posner and Herstein's pioneer results are very significant and a key factor for the study of noncommutative algebraic frames. Around 1950, Herstein first discovered the theory on Jordan and Lie ideals in simple, associative rings. He [9] also developed some theory on Lie ideals on rings of operators on a Hilbert space. In 1973, Awtar [10] first generalized the results of Posner in prime rings with the help of Lie and Jordan ideals. After that, Bergen et al. [11] discussed the connection between Lie ideal and derivation on the same structure, and finally measured the size of $\mathscr{D}(\Cal{U})$, where $\Cal{U}$ is a Lie ideal. They also highlighted the question of when $\mathscr{D}^2(\Cal{U})=\{0\}$ holds for a noncentral Lie ideal $\Cal{U}$ in a prime ring $\Cal{R}$. Later on, Martindale and Miers [12] worked on Lie ideals in semiprime rings with involution in 1986. %\? Lanski and Montgomery [13] discussed the Lie structures of prime rings of characteristics~$2$. %\?characteristic They proved that if $\Cal{U}$ is a commutative Lie ideal of a prime ring $\Cal{R}$, then $\Cal{U}\subset \Cal{Z}(\Cal{R})$ unless $\operatorname{char}(\Cal{R})= 2$ and $\Cal{R}\Cal{Z}^{-1}$ is a simple ring $4$-dimensional %\? over its center. Bre{\v{s}}ar proved in [14] that if $\mathscr{D}$ and $\mathscr{H}$ be two nonzero co-centralizing derivations on a prime ring $\Cal{R}$, then $\Cal{R}$ must be commutative. Later, Lee and Wong [15] studied similar conditions for noncentral Lie ideal %\?a $\Cal{L}$ of a prime ring $\Cal{R}$ and proved that if $\mathscr{D}(x)x-x\mathscr{H}(x)\in \Cal{Z}(\Cal{R})$ for all $x\in \Cal{L}$, then either $\mathscr{D}=O=\mathscr{H}$ or $\Cal{R}$ satisfies $\Cal{S}_4$. In [16], Lanski extended the Posner's result to the Lie ideal and proved that for a noncommutative Lie ideal $\Cal{L}$ of a prime ring $\Cal{R}$, if $[\mathscr{D}(x),x]_k=0$ for all $x\in \Cal{L}$ and $k>0$ fixed, then $\operatorname{char}(\Cal{R})=2$ and $\Cal{R}\subseteq \Cal{M}_2(\Cal{F})$ for a field $\Cal{F}$. After that, Carini [17] demonstrated that if $\Cal{L}$ is a noncentral Lie ideal of a prime ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$ and $\mathscr{D}$ is a nonzero derivation of~$\Cal{R}$ such that $[\mathscr{D}(u),u]^n\in \Cal{Z}(\Cal{R})$ for all $u\in \Cal{L}$ then $\Cal{R}$ satisfies the standard identity of degree $4$. Specifically, if $\mathscr{D}$ satisfies $[\mathscr{D}(u),u]^n=0$ for all $u\in \Cal{L}$ then $\Cal{R}$ is commutative. Following this development, Filippis [18] studied the same context, considering $\mathscr{D}$ as a~generalized derivation. Thereafter, Wang [19] investigated a similar scenario in which the derivation $\mathscr{D}$ is replaced with a nontrivial automorphism $\sigma$ of~$\Cal{R}$. He proved that if $\Cal{L}$ be a noncentral Lie ideal of a prime ring $\Cal{R}$ with center $\Cal{Z}(\Cal{R})$ and $\operatorname{char}(\Cal{R})$ either greater than $n$ or $0$, then for a nontrivial automorphism $\sigma$ of~$\Cal{R}$ such that $[\sigma(u),u]^n\in \Cal{Z}(\Cal{R})$ for all $u\in \Cal{L}$, $\Cal{R}$ satisfies $\Cal{S}_4$. Filippis and Huang [20] explored the action of skew derivations on Lie ideals and they showed in prime ring %\? $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$ and $3$ that if $\Cal{L}$ be a~noncentral Lie ideal of $\Cal{R}$, and $(\mathscr{D},\sigma)$ is a skew derivation of $\Cal{R}$ such that $[\mathscr{D}(x),x]^n=0$ for fixed positive integer $n$ and for all $x\in \Cal{L}$, then $\Cal{R}$ satisfies $\Cal{S}_4$. Later, Asma et al. [21] proved that if $\Cal{L}$ is a~noncentral Lie ideal of $\Cal{R}$ such that $u^2\in \Cal{L}$ for all $u\in \Cal{L}$ (that means, %\?много есть $\Cal{L}$ is a square closed noncentral Lie ideal) and $\mathscr{D}$ acts as a~homomorphism or antihomomorphism on $\Cal{L}$, then $\mathscr{D}=O$. Meanwhile, Wang and You [22] omitted the assumption $u^2\in \Cal{L}$ and examined the same results. Golbasi and Kaya [23] further showed that if the generalized derivation $\mathscr{G}$ of a prime ring $\Cal{R}$ acts as a homomorphism or antihomomorphism on a square closed Lie ideal $\Cal{L}$, then either $\mathscr{D}=O$ or $\Cal{L}$ is central in $\Cal{R}$. More detailed literature on Lie ideals can be found in the references [24--28] for further study. From the well-established and extensive literature on Lie ideals in some known algebraic structures, we try to present an investigation on the structure of Lie ideals in $rp$-algebra by using $d$-derivation $\Cal{D}_d$. This work is a continuation of our previous work [29]. In this article, we introduced the concept of $d$-derivation on an associative algebra over a ring and a prime structure, namely, $rp$-algebra. We also derived several differential identities related to commutativity for $rp$-algebra. Some important investigating identities are as follows: \Item (1) $\Cal{D}_{d}(xy)-xy \in \Cal{Z}(\Cal{A})$, \Item (2) $\Cal{D}_{d}([x,y]) + \Cal{D}_{d}(x)\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A})$, \Item (3) $[\Cal{D}_{d}(x), \Cal{D}_{d}(y)+y] \in \Cal{Z}(\Cal{A})$, \Item (4) $\Cal{D}_{d}(x\circ y) + \Cal{D}_{d}(x)\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A})$, \Item (5) $\Cal{D}_{d}([\Cal{D}_{d}(x),xy]) + x^{2}\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A})$ %$$ %\? так было в оригинале %\align % {1.}\ & \Cal{D}_{d}(xy)-xy \in \Cal{Z}(\Cal{A}),\\% % { 2.}\ & \Cal{D}_{d}([x,y]) + \Cal{D}_{d}(x)\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A}),\\ % { 3.}\ & [\Cal{D}_{d}(x), \Cal{D}_{d}(y)+y] \in \Cal{Z}(\Cal{A}),\\ % { 4.}\ & \Cal{D}_{d}(x\circ y) + \Cal{D}_{d}(x)\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A}),\\ % { 5.}\ & \Cal{D}_{d}([\Cal{D}_{d}(x),xy]) + x^{2}\Cal{D}_{d}(y) \in \Cal{Z}(\Cal{A}), %\endalign %$$ for all $x,y\in\Cal{A}$. In this paper, our main intention is to focus on determining the necessary and sufficient identities in connection with commutativity of the Lie ideals in $rp$-algebra. We delightedly establish the relationship between the center $\Cal{Z}(\Cal{A})$ and the centralizers of $\Cal{U}$ and $\Cal{D}_d(\Cal{U})$. We also construct certain identities by adopting the concepts of commuting and centralizing mappings. Furthermore, we categorize the Lie ideals in $rp$-algebra by restricting $d$-derivation $\Cal{D}_d$ to $\Cal{U}$ and considering its behavior as a homomorphism, antihomomorphism and Jordan homomorphism. Finally, we end our study by analyzing Posner and Herstein's results in $rp$-algebra using Lie ideals. \head 2. Preliminaries \endhead This section outlines some fundamental facts related to $d$-derivation and $rp$-algebra, which we proposed in [29]. \demo{Definition 2.1} Let $\Cal{A}$ be an associative algebra over a ring $\Cal{R}$. An additive map $\Cal{D}_d:\Cal{A}\to\Cal{A}$ is a~$d$-derivation if there exists a nonzero ring derivation $d:\Cal{R}\to\Cal{R}$ such that $$ \Cal{D}_d(ra) = d(r)a+r\Cal{D}_d(a)\ \text{ and }\ \Cal{D}_d(a_1a_2) = \Cal{D}_d(a_1)a_2 + a_1 \Cal{D}_d(a_2) $$ for all $a_1,a_2,a\in \Cal{A}$ and $r\in \Cal{R}$. \enddemo \demo{Definition 2.2} An associative algebra $\Cal{A}$ over a ring $\Cal{R}$ is an $rp$-algebra if $\Cal{A}$ does not contain divisors of zero and for any $r\in \Cal{R}$ and $p\in \Cal{A}$, $r\Cal{A}p=\{0\}$ implies either $r=0$ or $p=0$. \enddemo \proclaim{Lemma 2.3} Let $\Cal{A}$ be an $rp$-algebra over a ring $\Cal{R}$. For any $r(\neq 0)\in \Cal{R}$ and $a, b\in \Cal{A}$, the following holds: %\?hold \Item (i) $ra=0 \Rightarrow a=0$. \Item (ii) $ra \in \Cal{Z}(\Cal{A}) \Rightarrow a \in \Cal{Z}(\Cal{A})$. \Item (iii) If $b\;(\neq 0)\in \Cal{Z}(\Cal{A})$ and $ab\in \Cal{Z}(\Cal{A})$ {\rm(}or $ba\in \Cal{Z}(\Cal{A})${\rm)} then $a\in \Cal{Z}(\Cal{A})$. \endproclaim \proclaim{Theorem 2.4} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. If $[\Cal{D}_d(x),x]=0$ for all $x\in \Cal{A}$, then $\Cal{A}$ is commutative. \endproclaim \proclaim{Lemma 2.5} Let $\Cal{A}$ be an $rp$-algebra over a ring $\Cal{R}$ and let $\Cal{D}_d$ be a $d$-derivation on $\Cal{A}$. If $x\circ y = 0$ for all $x,y\in \Cal{A}$, then $\Cal{A}$ is commutative. \endproclaim \proclaim{Theorem 2.6} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. If $x\circ y \in \Cal{Z}(\Cal{A})$ for all $x,y\in \Cal{A}$, then $\Cal{A}$ is commutative. \endproclaim \proclaim{Theorem 2.7} An $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ admitting a $d$-derivation $\Cal{D}_d$ is commutative if $\Cal{D}_d(x)\circ x \in \Cal{Z}(\Cal{A})$ for all $x\in \Cal{A}$. \endproclaim \proclaim{Theorem 2.8} Let an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R})\neq n$ admits %\?admit a nonzero $d$-derivation $\Cal{D}_d$ such that $d$ is commuting on $\Cal{R}$. If $\Cal{D}_d(x^n)\in \Cal{Z}(\Cal{A})$ for all $x\in \Cal{A}$, then $\Cal{A}$ is commutative. \endproclaim \proclaim{Theorem 2.9} Let $\Cal{D}_{d_1}$ and $\Cal{D}_{d_2}$ be respectively $d_1$-derivation and $d_2$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$ such that $\Cal{D}_{d_1}\Cal{D}_{d_2}$ is a $d_1d_2$-derivation on $\Cal{A}$. Then either $\Cal{D}_{d_1}=O$ or $\Cal{D}_{d_2}=O$. \endproclaim Throughout the paper, we use $\Cal{A}$ as an $rp$-algebra over a ring with unity $\Cal{R}$ (not necessarily commutative), $\Cal{Z}(\Cal{A})$ as the center of $\Cal{A}$ and $\Cal{Q}=\{r \in \Cal{R} : d(r) \neq 0\}$. By $C_{\Cal{A}}(\Cal{U})$, we will mean the centralizer of~$\Cal{U}$, defined by $C_{\Cal{A}}(\Cal{U}) = \{x \in \Cal{A} : xu = ux \text{ for all } u \in \Cal{U}\}$. \head 3. $d$-Derivation on Lie Ideals of $rp$-Algebras \endhead We start our investigation with the following results, which will be referred to throughout our study. \proclaim{Lemma 3.1} Let $\Cal{U}$ be a Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. For some $r \in \Cal{R}$ and $a \in \Cal{A}$, if $r \Cal{U} a = \{0\}$, then either $r = 0$ or $a = 0$. \endproclaim \demo{Proof} Suppose $r \Cal{U} a = \{0\}$ for some $r \in \Cal{R}$ and $a \in \Cal{A}$. Pre-multiplying %\? both sides by $b$, we get $$ r (b\,\Cal{U}a) = \{0\} \text{ for all } b\in \Cal{A} \Rightarrow r \Cal{A}\Cal{U} a = \{0\}. $$ Since $\Cal{A}$ is an $rp$-algebra and $\Cal{U} \neq \{0\}$, it easily follows that either $r = 0$ or $a = 0$. \proclaim{Lemma 3.2} For a Lie ideal $\Cal{U}$ of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$, if $\Cal{D}_d(\Cal{U}) = \{0\}$ then $$ \Cal{D}_d(\Cal{A}) \subseteq C_\Cal{A}(\Cal{U}). $$ \endproclaim \demo{Proof} Suppose $\Cal{D}_d(\Cal{U}) = \{0\}$, which means $$ \Cal{D}_d(u) = 0 \quad \text{for all } u \in \Cal{U}. \tag3.1 $$ Now, replacing $u$ by $[u,a]$ for all $a \in \Cal{A}$, we have $\Cal{D}_d([u,a]) = 0$. It implies that $$ [\Cal{D}_d(u),a] + [u, \Cal{D}_d(a)] = 0 \Rightarrow [u, \Cal{D}_d(a)] = 0 \quad \text{(using (3.1))}. $$ Therefore, $\Cal{D}_d(a) \in C_\Cal{A}(\Cal{U})$ for all $a \in \Cal{A}$ and hence, $\Cal{D}_d(\Cal{A}) \subseteq C_\Cal{A}(\Cal{U})$. \proclaim{Lemma 3.3} If $\Cal{U}$ is a noncentral Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$, then $$ C_\Cal{A}([\Cal{U},\Cal{A}]) = C_\Cal{A}(\Cal{U}). $$ \endproclaim \demo{Proof} Let $x \in C_\Cal{A}([\Cal{U},\Cal{A}])$. It implies that $$ [x,[u,a]] = 0 \quad \text{for all } u \in \Cal{U} \text{ and } a \in \Cal{A}. \tag3.2 $$ Now, putting $a$ by $au$, we get $[x,[u,au]] =0$. Therefore, $$ [x,[u,a] u] = 0, \text{ which implies } [u,a][x,u] = 0 \quad \text{(using (3.2))}. $$ Since $\Cal{U} \nsubseteq \Cal{Z}(\Cal{A})$ and $\Cal{A}$ is an $rp$-algebra, it follows that $[x,u] = 0$ for all $u \in \Cal{U}$. Hence $x \in C_\Cal{A}(\Cal{U})$, which implies $C_\Cal{A}([\Cal{U},\Cal{A}]) \subseteq C_\Cal{A}(\Cal{U})$. Conversely, let $y \in C_{\Cal{A}}(\Cal{U})$, that is, $[y,u]=0$ for all $u \in \Cal{U}$. Replacing $u$ by $[u,a]$ for all $a \in \Cal{A}$, we obtain $[y,[u,a]] = 0$ and hence, $y \in C_\Cal{A}([\Cal{U},\Cal{A}])$. Thus, $C_\Cal{A}(\Cal{U}) \subseteq C_\Cal{A}([\Cal{U},\Cal{A}])$, and the equality follows. \proclaim{Lemma 3.4} In an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$, if $\Cal{U}$ is a noncentral Lie ideal of~$\Cal{A}$ such that $$ [\Cal{U},\Cal{U}] \subseteq C_\Cal{A}(\Cal{U}), $$ then $\Cal{U}$ is commutative. \endproclaim \demo{Proof} For all $u \in \Cal{U}$ and $a \in \Cal{A}$, we have $[u,a] \in \Cal{U}$. Let us consider $ x = [u,[u,a]] \in [\Cal{U},\Cal{U}] \subseteq C_\Cal{A}(\Cal{U})$. Then $$ ux = u[u,[u,a]] = [u,[u,ua]] \in C_\Cal{A}(\Cal{U}). $$ Therefore $[ux,v] =0$ for all $v \in \Cal{U}$. It implies that $$ [u,v]x + u[x,v] = 0\Rightarrow [u,v]x = 0. $$ {\sc Case 1:} If $x \neq 0$ then $[u,v] = 0$ for all $u,v \in \Cal{U}$ and hence, $\Cal{U}$ is commutative. {\sc Case 2:} If $x = 0$, that is, $[u,[u,a]] = 0$ for all $u \in \Cal{U}$. Replacing $u$ by $u+v$ for all $v \in \Cal{U}$, we get $$ [u+v,[u+v,a]] = 0 \Rightarrow [u,[v,a]] + [v,[u,a]] = 0. \tag3.3 $$ Again, using the Jacobi identity, we can write $$ [u,[v,a]] + [v,[a,u]] + [a,[u,v]] = 0, \text{ that means, } [u,[v,a]] - [v,[u,a]] + [a,[u,v]] = 0. \tag3.4 $$ Combining \Tag(3.3) and \Tag(3.4), we get $$ [a,[u,v]] = -2 [u,[v,a]] \in C_\Cal{A}(\Cal{U}). \tag3.5 $$ Take $c = [u,v]$. Then $[a,c] \in C_\Cal{A}(\Cal{U})$, which implies $[u,[a,c]] = 0$. Replacing $[a,c]$ by $v[a,c]$ for all $v \in \Cal{U}$, we obtain $$ [u,v[a,c]] = 0 \text{ implies } [u,v][a,c] = 0. $$ Thus it follows that either $[u,v]=0$ or $[a,c]=0$. If $[u,v]=0$ for all $u,v\in \Cal{U}$, then $\Cal{U}$ is commutative. But, if $[a,c] = 0$, then by \Tag(3.5) we have $2 [u,[v,a]] = 0$. Applying the $d$-derivation $\Cal{D}_d$ and expanding, we get $$ 2 ([\Cal{D}_d(u),[v,a]] + [u,[\Cal{D}_d(v),a]] + [u,[v,\Cal{D}_d(a)]] ) = 0. \tag3.6 $$ Finally, substituting $a$ by $ra$, where $r \in \Cal{Q}$ and using $\operatorname{char}(\Cal{R}) \neq 2$, we obtain $$ 2d(r)[u,[v,a]] = 0 \Rightarrow [u,[v,a]] = 0. $$ It implies that $u \in C_\Cal{A}([\Cal{U},\Cal{A}])$. So, by \Par*{Lemma 3.3}, we obtain $u \in C_\Cal{A}(\Cal{U})$ for all $u\in \Cal{U}$, that means, $\Cal{U} \subseteq C_\Cal{A}(\Cal{U})$. Thus for all cases, $\Cal{U}$ is commutative. \head 4. Main Results \endhead In the main results, we first discuss the behavior of the image of a Lie ideal $\Cal{U}$ under $\Cal{D}_d$. \proclaim{Theorem 4.1} If a Lie ideal $\Cal{U}$ of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ be contained in the image $\Cal{D}_d(\Cal{U})$, then $\Cal{D}_d(\Cal{U})$ is a Lie ideal of $\Cal{A}$. \endproclaim \demo{Proof} First, we claim that $\Cal{D}_d(\Cal{U})$ is an $\Cal{R}$-submodule of $\Cal{A}$. Consider $r \in \Cal{R}$ and $\Cal{D}_d(u), \Cal{D}_d(v) \in \Cal{D}_d(\Cal{U})$, where $u,v \in \Cal{U}$. Then $$ \Cal{D}_d(u) + \Cal{D}_d(v) = \Cal{D}_d(u+v) \in \Cal{D}_d(\Cal{U}) \text{ and } \Cal{D}_d(ru) = d(r)u + r \Cal{D}_d(u) \in \Cal{D}_d(\Cal{U}). $$ Since $ru, d(r)u \in \Cal{U} \subseteq \Cal{D}_d(\Cal{U})$; $r\Cal{D}_d(u)\in \Cal{D}_d(\Cal{U})$ and hence, $\Cal{D}_d(\Cal{U})$ is an $\Cal{R}$-submodule of $\Cal{A}$. To prove $\Cal{D}_d(\Cal{U})$ as a Lie ideal, we have to show that $[\Cal{D}_d(u),a]\in \Cal{D}_d(\Cal{U})$ for all $a\in \Cal{A}$. For Lie ideal, $[u,a] \in \Cal{U}$. Then $$ \Cal{D}_d([u,a]) \in \Cal{D}_d(\Cal{U}),\quad \text{which gives } [\Cal{D}_d(u), a] + [u, \Cal{D}_d(a)] \in \Cal{D}_d(\Cal{U}). $$ But, $[u, \Cal{D}_d(a)] \in \Cal{D}_d(\Cal{U})$. Hence, we conclude that $$ [\Cal{D}_d(u), a] \in \Cal{D}_d(\Cal{U}). $$ Therefore, $\Cal{D}_d(\Cal{U})$ is a Lie ideal of $\Cal{A}$. \proclaim{Theorem 4.2} If $\Cal{U}$ is a Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$ which admits a $d$-derivation $\Cal{D}_d$ such that $$ {\Cal{D}_d}^2(\Cal{U}) = \{0\}, $$ then $\Cal{U} \subseteq \Cal{Z}(\Cal{A})$. \endproclaim \demo{Proof} Suppose ${\Cal{D}_d}^2(\Cal{U}) = \{0\}$, that is, $$ {\Cal{D}_d}^2(u) = 0 \quad \text{for all } u \in \Cal{U}. \tag4.1 $$ Now, replacing $u$ by $[u,a]$ for all $a \in \Cal{A}$, we get ${\Cal{D}_d}^2([u,a]) = 0 $, which implies $$ \Cal{D}_d([\Cal{D}_d(u),a] + [u,\Cal{D}_d(a)]) = 0. $$ Again expanding, $$ [{\Cal{D}_d}^2(u),a] + 2[\Cal{D}_d(u),\Cal{D}_d(a)] + [u,{\Cal{D}_d}^2(a)] = 0. $$ Using \Tag(4.1), this reduces to $$ 2[\Cal{D}_d(u),\Cal{D}_d(a)] + [u,{\Cal{D}_d}^2(a)] = 0. \tag4.2 $$ Again, setting $a$ to $au$, we have $$ \align 2[\Cal{D}_d(u),& \Cal{D}_d(au)] + [u,{\Cal{D}_d}^2(au)] = 0 \\ & \Rightarrow 2[\Cal{D}_d(u),\Cal{D}_d(a)]u+2\Cal{D}_d(a)[\Cal{D}_d(u),u]+2[\Cal{D}_d(u),a]\Cal{D}_d(u)+[u,{\Cal{D}_d}^2(a)]u+2[u,\Cal{D}_d(a)]\Cal{D}_d(u) \\ &\qquad +2\Cal{D}_d(a)[u,\Cal{D}_d(u)] +[u,a]{\Cal{D}_d}^2(u)+a[u,{\Cal{D}_d}^2(u)]=0. \endalign $$ Using \Tag(4.1) and \Tag(4.2), we get $$ 2[\Cal{D}_d(u),a]\Cal{D}_d(u) + 2[u,\Cal{D}_d(a)]\Cal{D}_d(u) = 0. \tag4.3 $$ Replacing $a$ by $ra$, where $r \in \Cal{Q}$, we obtain $$ \align 2[\Cal{D}_d(u),ra]&\Cal{D}_d(u) + 2[u,\Cal{D}_d(ra)]\Cal{D}_d(u) = 0 \\ &\Rightarrow 2r[\Cal{D}_d(u),a]\Cal{D}_d(u) + 2 d(r)[u,a]\Cal{D}_d(u) + 2r[u,\Cal{D}_d(a)]\Cal{D}_d(u) = 0. \endalign $$ Using \Tag(4.3), it follows that $$ 2 d(r)[u,a]\Cal{D}_d(u) = 0 \text{ that means } [u,a]\Cal{D}_d(u) = 0 \quad \text{(since $\operatorname{char}(\Cal{R}) \neq 2$)}. \tag4.4 $$ Setting $u$ by $u+v$ for all $v \in \Cal{U}$, we have $$ [u,a]\Cal{D}_d(u) + [u,a]\Cal{D}_d(v) + [v,a]\Cal{D}_d(u) + [v,a]\Cal{D}_d(v) = 0. $$ Applying equation \Tag(4.4), we deduce $$ [u,a]\Cal{D}_d(v) + [v,a]\Cal{D}_d(u) = 0. \tag4.5 $$ Finally, substituting $v$ by $r_1v$, where $r_1 \in \Cal{Q}$, we obtain $$ [u,a]\Cal{D}_d(r_1v) + [r_1v,a]\Cal{D}_d(u) = 0. $$ Expanding, $$ d(r_1)[u,a]v + r_1[u,a]\Cal{D}_d(v) + r_1[v,a]\Cal{D}_d(u) = 0. $$ Lastly, for $d(r_1) \neq 0$ and using \Tag(4.5), we get $ [u,a]v = 0$ for all $v \in \Cal{U}$. Since $\Cal{U}$ is a nonzero Lie ideal of~$\Cal{A}$, it follows that $[u,a] = 0$ for all $u \in \Cal{U}$ and $a \in \Cal{A}$. Hence, $ \Cal{U} \subseteq \Cal{Z}(\Cal{A})$. \proclaim{Corollary 4.3} If $\Cal{U}$ is a Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$, which admits a $d$-derivation $\Cal{D}_d$ such that ${\Cal{D}_d}(u)\not= 0$ for all $u(\neq 0)\in \Cal{U}$, but ${\Cal{D}_d}^2(\Cal{U})=\{0\}$ then $\Cal{U}$ is central on $\Cal{A}$. \endproclaim \proclaim{Theorem 4.4} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. If all elements of a Lie ideal $\Cal{U}$ of $\Cal{A}$ are idempotent, then $\Cal{U}$ is central on $\Cal{A}$. \endproclaim \demo{Proof} For all $u \in \Cal{U}$, we have $ \Cal{D}_d(u^2) = \Cal{D}_d(u)$. Replacing $u$ by $[u,a]$ for all $a \in \Cal{A}$, we obtain $ \Cal{D}_d([u,a]^2) = \Cal{D}_d([u,a])$. Expanding the left-hand side, we get $$ \align \Cal{D}_d&([u,a])[u,a] + [u,a] \Cal{D}_d([u,a]) = \Cal{D}_d([u,a]) \\ &\Rightarrow [\Cal{D}_d(u),a][u,a] + [u,\Cal{D}_d(a)][u,a] + [u,a][\Cal{D}_d(u),a] + [u,a][u,\Cal{D}_d(a)] = [\Cal{D}_d(u),a] + [u,\Cal{D}_d(a)]. \endalign $$ Now, pre-multiplying both sides by $[u,a]$, we get $$ \align [u,a]&[\Cal{D}_d(u),a][u,a] + [u,a][u,\Cal{D}_d(a)][u,a] + [u,a]^2[\Cal{D}_d(u),a] + [u,a]^2[u,\Cal{D}_d(a)] \\ &= [u,a][\Cal{D}_d(u),a] + [u,a][u,\Cal{D}_d(a)] \\ &\Rightarrow [u,a][\Cal{D}_d(u),a][u,a]+[u,a][u,\Cal{D}_d(a)][u,a]+[u,a][\Cal{D}_d(u),a]+[u,a][u,\Cal{D}_d(a)] \\ &=[u,a][\Cal{D}_d(u),a]+[u,a][u,\Cal{D}_d(a)]. \endalign $$ This implies that $$ [u,a][\Cal{D}_d(u),a][u,a] + [u,a][u,\Cal{D}_d(a)][u,a] = 0, $$ which can be written as $$ [u,a] \Cal{D}_d([u,a]) [u,a] = 0. $$ Therefore, if $[u,a] = 0$, then we are done. But if $\Cal{D}_d([u,a]) = 0$, then substituting $u$ by $ru$, where $r \in \Cal{Q}$, we get $$ d(r)[u,a] + r \Cal{D}_d([u,a]) = 0 \Rightarrow d(r)[u,a] = 0, $$ which gives $[u,a] = 0$ for all $u \in \Cal{U}$ and $a \in \Cal{A}$. Hence $ \Cal{U} \subseteq \Cal{Z}(\Cal{A})$, that means, $\Cal{U}$ is central on $\Cal{A}$. \proclaim{Theorem 4.5} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ such that ring derivation $d$ is not commuting on $\Cal{R}$. Then a noncentral Lie ideal $\Cal{U}$ of $\Cal{A}$ is commutative if and only if $$ [\Cal{D}_d(u),u]=0 \quad \text{for all $u\in \Cal{U}$}. $$ \endproclaim \demo{Proof} Let $[\Cal{D}_d(u),u]=0$ for all $u\in \Cal{U}$. Now, substitute $u$ by $u+v$ for all $v\in\Cal{U}$, we obtain $[\Cal{D}_d(u+v),u+v]=0$, which gives $$ [\Cal{D}_d(u),v]+[\Cal{D}_d(v),u]=0. \tag4.6 $$ Again, replacing $u$ by $ru$ for any $r\in \Cal{Q}$, we have $$ [\Cal{D}_d(ru),v]+[\Cal{D}_d(v),ru]=0 \Rightarrow d(r)[u,v]+r([\Cal{D}_d(u),v]+[\Cal{D}_d(v),u])=0. $$ Using \Tag(4.6), we get $d(r)[u,v]=0$. This implies that $[u,v]=0$ for all $u,v\in\Cal{U}$. Hence $\Cal{U}$ is commutative. Conversely, assume that $\Cal{U}$ is commutative, that means, $[u,v]=0$ for all $u,v\in \Cal{U}$. Therefore, $$ \Cal{D}_d([u,v])=0,\quad \text{which implies } [\Cal{D}_d(u),v]+[u,\Cal{D}_d(v)]=0. \tag4.7 $$ Now, we replace $u$ by $[u,a]$ for all $a\in\Cal{A}$, $$ \aligned [\Cal{D}_d([u,a]),&v] +[[u,a],\Cal{D}_d(v)]=0 \\ & \Rightarrow [[\Cal{D}_d(u),a],v]+[[u,\Cal{D}_d(a)],v]+[[u,a],\Cal{D}_d(v)]=0 \\ & \Rightarrow [[\Cal{D}_d(u),a],v]+[[u,a],\Cal{D}_d(v)]=0. \endaligned \tag4.8 $$ Again, substitute $a$ by $a\Cal{D}_d(u)$, we obtain $$ \align [[\Cal{D}_d(u),&a\Cal{D}_d(u)],v] + [[u,a\Cal{D}_d(u)],\Cal{D}_d(v)] =0 \\ & \Rightarrow [[\Cal{D}_d(u),a]\Cal{D}_d(u),v] +[a[\Cal{D}_d(u),\Cal{D}_d(u)],v] +[\,[u,a]\Cal{D}_d(u),\Cal{D}_d(v)] +[a[u,\Cal{D}_d(u)],\Cal{D}_d(v)] =0 \\ & \Rightarrow [\Cal{D}_d(u),a][\Cal{D}_d(u),v]+[[\Cal{D}_d(u),a],v]\Cal{D}_d(u)+[u,a][\Cal{D}_d(u),\Cal{D}_d(v)]+[[u,a],\Cal{D}_d(v)]\Cal{D}_d(u) \\ &\qquad+a[[u,\Cal{D}_d(u)],\Cal{D}_d(v)]+[a,\Cal{D}_d(v)][u,\Cal{D}_d(u)]=0 \\ &\Rightarrow [\Cal{D}_d(u),a][\Cal{D}_d(u),v] +[u,a][\Cal{D}_d(u),\Cal{D}_d(v)] +a[[u,\Cal{D}_d(u)],\Cal{D}_d(v)] +[a,\Cal{D}_d(v)][u,\Cal{D}_d(u)] =0. \endalign $$ Also, replacing $u$ by $ru$ for any $r\in\Cal{Q}$, $$ [\Cal{D}_d(ru),a][\Cal{D}_d(ru),v] +[ru,a][\Cal{D}_d(ru),\Cal{D}_d(v)] +a[[ru,\Cal{D}_d(ru)],\Cal{D}_d(v)] +[a,\Cal{D}_d(v)][ru,\Cal{D}_d(ru)] =0. $$ This simplifies to $$ d(r)r[u,a][\Cal{D}_d(u),v] + rd(r)[u,a][u,\Cal{D}_d(v)] =0. $$ Finally, substituting $v$ by $u$, we get $$ \align d(r)r&[u,a][\Cal{D}_d(u),u] + rd(r)[u,a][u,\Cal{D}_d(u)] =0 \\ & \Rightarrow d(r)r[u,a][\Cal{D}_d(u),u] - rd(r)[u,a][\Cal{D}_d(u),u] =0 \\ & \Rightarrow [d(r),r][u,a][\Cal{D}_d(u),u]=0. \endalign $$ Since $\Cal{U}\nsubseteq \Cal{Z}(\Cal{A})$ and $[d(r),r]\neq 0$, we conclude that $[\Cal{D}_d(u),u]=0$ for all $u\in\Cal{U}$. \proclaim{Theorem 4.6} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$. Then a noncentral Lie ideal $\Cal{U}$ of $\Cal{A}$ is commutative if and only if $$ [\Cal{D}_d(u), u] \in C_\Cal{A}(\Cal{U})\quad\text{for all }u \in \Cal{U}. $$ \endproclaim \demo{Proof} Suppose $[\Cal{D}_{d}(u),u]\in C_{\Cal{A}}(\Cal{U})$ for all $u\in \Cal{U}$. Replacing $u$ by $u+v$ for all $v\in \Cal{U}$, we obtain $$ [\Cal{D}_{d}(u+v),u+v]\in C_{\Cal{A}}(\Cal{U}) \Rightarrow [\Cal{D}_{d}(u),v]+[\Cal{D}_{d}(v),u] \in C_{\Cal{A}}(\Cal{U}). \tag4.9 $$ Substitute $u$ by $ru$ for any $r\in \Cal{Q}$, we have $$ [\Cal{D}_{d}(ru),v]+[\Cal{D}_{d}(v),ru] \in C_{\Cal{A}}(\Cal{U}) \Rightarrow d(r)[u,v]+r([\Cal{D}_{d}(u),v]+[\Cal{D}_{d}(v),u]) \in C_{\Cal{A}}(\Cal{U}). $$ Using \Tag(4.9), it follows that $ d(r)[u,v]\in C_{\Cal{A}}(\Cal{U})$ and consequently, $[u,v]\in C_{\Cal{A}}(\Cal{U})$ for all $u,v\in \Cal{U}$, that means, $[\Cal{U},\Cal{U}] \subseteq C_{\Cal{A}}(\Cal{U})$. Hence by \Par*{Lemma~3.4}, we conclude that $\Cal{U}$ is commutative. Conversely, let $\Cal{U}$ be commutative, that is, $[u,v]=0$ for all $u,v\in \Cal{U}$. Therefore, $$ [[u,\Cal{D}_{d}(u)],v]=0 \quad\text{for all } u,v\in \Cal{U} $$ and as a result, we obtain $[u,\Cal{D}_d(u)]\in C_\Cal{A}(\Cal{U})$. \proclaim{Theorem 4.7} In an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ that admits a $d$-derivation $\Cal{D}_d$, if $\Cal{U}$ is a Lie ideal of $\Cal{A}$ such that $\Cal{D}_d(\Cal{U})$ is commutative, then $[\Cal{U},\Cal{U}]$ is central on $\Cal{A}$. \endproclaim \demo{Proof} Consider $ [\Cal{D}_d(u),\Cal{D}_d(v)] = 0$ for all $ u,v \in \Cal{U}$. Replacing $u$ by $[u,a]$ for all $a \in \Cal{A}$, we have $$ [\Cal{D}_d([u,a]),\Cal{D}_d(v)] = 0,\quad\text{which implies } [\Cal{D}_d(ua - au), \Cal{D}_d(v)] = 0. $$ This gives $$ [\Cal{D}_d(ua), \Cal{D}_d(v)] - [\Cal{D}_d(au), \Cal{D}_d(v)] = 0. $$ Expanding, we get $$ \align [\Cal{D}_d(u),\Cal{D}_d(v)] a &+ \Cal{D}_d(u) [a,\Cal{D}_d(v)] + [u,\Cal{D}_d(v)] \Cal{D}_d(a) + u [\Cal{D}_d(a),\Cal{D}_d(v)]- [\Cal{D}_d(a),\Cal{D}_d(v)] u \\ &- \Cal{D}_d(a) [u,\Cal{D}_d(v)] - [a,\Cal{D}_d(v)] \Cal{D}_d(u) - a [\Cal{D}_d(u),\Cal{D}_d(v)] = 0. \endalign $$ After simplification, $$ [\Cal{D}_d(u), [a,\Cal{D}_d(v)]] + [[u, \Cal{D}_d(v)], \Cal{D}_d(a)] + [u, [\Cal{D}_d(a), \Cal{D}_d(v)]] = 0. \tag4.10 $$ Now, putting $u = ru$, where $r \in \Cal{Q}$, we get $$ d(r)[u,[a,\Cal{D}_d(v)]] + r ( [\Cal{D}_d(u), [a, \Cal{D}_d(v)]] + [[u, \Cal{D}_d(v)], \Cal{D}_d(a)] + [u, [\Cal{D}_d(a), \Cal{D}_d(v)]] ) = 0. $$ Using \Tag(4.10) and $d(r) \neq 0$, we obtain $$ [u,[a,\Cal{D}_d(v)]] = 0. \tag4.11 $$ Again, setting $a$ by $r_1 a$ for any $r_1 \in \Cal{Q}$ in \Tag(4.10), we get $$ d(r_1) ( [[u, \Cal{D}_d(v)], a] + [u, [a, \Cal{D}_d(v)]] ) + r_1 ( [\Cal{D}_d(u), [a, \Cal{D}_d(v)]] + [[u, \Cal{D}_d(v)], \Cal{D}_d(a)] + [u, [\Cal{D}_d(a), \Cal{D}_d(v)]] ) = 0. $$ Using \Tag(4.10), this reduces to $$ d(r_1) ( [[u, \Cal{D}_d(v)], a] + [u, [a, \Cal{D}_d(v)]] ) = 0\;\Rightarrow \; [[u, \Cal{D}_d(v)], a] + [u, [a, \Cal{D}_d(v)]] = 0. $$ Also, using \Tag(4.11), we get $$ [[u, \Cal{D}_d(v)], a] = 0. $$ Finally, substituting $v = r_2 v$ for any $r_2 \in \Cal{Q}$, we find $$ d(r_2) [[u,v],a] + r_2 [[u, \Cal{D}_d(v)], a] = 0,\quad\text{which reduces }[[u,v],a] = 0. $$ This implies that $[u,v] \in \Cal{Z}(\Cal{A})$ for all $u,v \in \Cal{U}$, that means, $[\Cal{U}, \Cal{U}] \subseteq \Cal{Z}(\Cal{A})$. Hence $[\Cal{U}, \Cal{U}]$ is central on $\Cal{A}$. From the above theorem, we immediately conclude an important remark. \demo{Remark 4.8} In an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ which admits a $d$-derivation $\Cal{D}_d$, if $\Cal{U}$ is a Lie ideal of $\Cal{A}$ such that $\Cal{D}_d(\Cal{U})$ is commutative then $\Cal{U}$ may not be central. But there exists a Lie ideal of the form $[\Cal{U},\Cal{U}]$ which is central. \enddemo \proclaim{Theorem 4.9} Let $\Cal{D}_d$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$. Then a noncentral Lie ideal $\Cal{U}$ of $\Cal{A}$ is commutative if and only if $$ [\Cal{D}_d(\Cal{U}), \Cal{D}_d(\Cal{U})] \subseteq C_\Cal{A}(\Cal{U}). $$ \endproclaim \demo{Proof} Suppose $$ [\Cal{D}_d(u), \Cal{D}_d(v)] \in C_\Cal{A}(\Cal{U})\quad\text{for all }u,v \in \Cal{U}. \tag4.12 $$ Replacing $u$ by $ru$ for any $r \in \Cal{Q}$, we obtain $$ [\Cal{D}_d(ru), \Cal{D}_d(v)] \in C_\Cal{A}(\Cal{U}) \Rightarrow d(r)[u, \Cal{D}_d(v)] + r[\Cal{D}_d(u), \Cal{D}_d(v)] \in C_\Cal{A}(\Cal{U}). $$ Using \Tag(4.12) and $d(r)\neq 0$, we get $ [u, \Cal{D}_d(v)] \in C_\Cal{A}(\Cal{U})$. Again, substituting $v$ by $r_1 v$, where $r_1 \in \Cal{Q}$, $$ d(r_1)[u,v] + r_1 [u, \Cal{D}_d(v)] \in C_\Cal{A}(\Cal{U}). $$ It implies $[u,v] \in C_\Cal{A}(\Cal{U})$ for all $u,v\in \Cal{U}$, that means, $[\Cal{U},\Cal{U}]\subseteq C_\Cal{A}(\Cal{U})$. So, \Par*{Lemma 3.4} implicates that $\Cal{U}$ is commutative. Conversely, suppose $\Cal{U}$ is commutative. Then \Par*{Theorem 4.5} implies $[\Cal{D}_d(u), u] = 0$ for all $u \in \Cal{U}$. Putting $u$ by $u+v$ and expanding, we obtain $ [\Cal{D}_d(u), v] + [\Cal{D}_d(v), u] = 0$. Now, applying $\Cal{D}_d$ on both sides, we get $$ [{\Cal{D}_d}^2(u),v]+[\Cal{D}_d(u),\Cal{D}_d(v)]+[\Cal{D}_d(v),\Cal{D}_d(u)]+[{\Cal{D}_d}^2(v),u]=0, $$ which reduces to $$ [{\Cal{D}_d}^2(u), v] + [{\Cal{D}_d}^2(v), u] = 0. \tag4.13 $$ Replacing $u$ by $r_2u$, where $r_2\in \Cal{Q}$, we obtain $$ \align [{\Cal{D}_d}^2(r_2u),v] &+[{\Cal{D}_d}^2(v),r_2u]=0 \\ &\Rightarrow d^2(r_2)[u,v]+2d(r_2)[\Cal{D}_d(u),v]+r_2([{\Cal{D}_d}^2(u),v]+[{\Cal{D}_d}^2(v),u])=0 \\ & \Rightarrow 2d(r_2)[\Cal{D}_d(u),v]=0\quad\text{(using (4.13))}. \endalign $$ Again, using $\operatorname{char}(\Cal{R}) \neq 2$, we have $ [\Cal{D}_d(u), v] = 0$. Finally, setting $v$ by $[v, \Cal{D}_d(v)]$, we find $$ \align [\Cal{D}_d(u),[v,\Cal{D}_d(v)]]=0 & \Rightarrow [\Cal{D}_d(u),v\Cal{D}_d(v)]-[\Cal{D}_d(u),\Cal{D}_d(v)v]=0 \\ & \Rightarrow v[\Cal{D}_d(u),\Cal{D}_d(v)]+[\Cal{D}_d(u),v]\Cal{D}_d(v)-\Cal{D}_d(v)[\Cal{D}_d(u),v]-[\Cal{D}_d(u),\Cal{D}_d(v)]v=0 \\ & \Rightarrow [v,[\Cal{D}_d(u),\Cal{D}_d(v)]]=0 \quad\text{for all }u,v\in \Cal{U}. \endalign $$ This implicates that $[\Cal{D}_d(u),\Cal{D}_d(v)]\in C_\Cal{A}(\Cal{U})$ for all $u,v\in \Cal{U}$ and hence, $ [\Cal{D}_d(\Cal{U}), \Cal{D}_d(\Cal{U})] \subseteq C_\Cal{A}(\Cal{U})$. \proclaim{Theorem 4.10}\Label{T4.10} Let $\Cal{U}$ be a Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ which admits a $d$-derivation $\Cal{D}_d$. Then \Item (i) $\Cal{U} \subseteq \Cal{Z}(\Cal{A})$ if and only if $\Cal{D}_d(\Cal{U}) \subseteq \Cal{Z}(\Cal{A})$. \Item (ii) $\Cal{U}\subseteq C_\Cal{A}(\Cal{U})$ if and only if $\Cal{D}_d(\Cal{U})\subseteq C_\Cal{A}(\Cal{U})$, where $\Cal{U}$ is noncentral and $d$ is not commuting on $\Cal{R}$. \endproclaim \demo{Proof} \Par{T4.10}{(i)}: Suppose $\Cal{U} \subseteq \Cal{Z}(\Cal{A})$, that is, $[u,a] = 0$ for all $u \in \Cal{U}$ and $a \in \Cal{A}$. Therefore $\Cal{D}_d([u,a]) = 0$, which implies $$ [\Cal{D}_d(u), a] + [u, \Cal{D}_d(a)] = 0,\text{ that means},\ [\Cal{D}_d(u), a] = 0\text{ for all }a \in \Cal{A}. $$ It concludes that $\Cal{D}_d(u) \in \Cal{Z}(\Cal{A})$ for all $u \in \Cal{U}$ and consequently, $\Cal{D}_d(\Cal{U}) \subseteq \Cal{Z}(\Cal{A})$. Conversely, suppose $\Cal{D}_d(\Cal{U}) \subseteq \Cal{Z}(\Cal{A})$, that means, $\Cal{D}_d(u) \in \Cal{Z}(\Cal{A})$ for all $u \in \Cal{U}$. This implies that $$ [\Cal{D}_d(u), a] = 0 \quad \text{for all } a \in \Cal{A}. \tag4.14 $$ Replacing $u$ by $ru$ for any $r \in \Cal{Q}$, we get $[\Cal{D}_d(ru), a] = 0$, which gives $$ d(r)[u,a] + r[\Cal{D}_d(u), a] = 0 \; \Rightarrow \; d(r)[u,a] = 0 \quad \text{(using (4.14))}. $$ Hence $ [u,a] = 0$ for all $a \in \Cal{A}$, that means, $u \in \Cal{Z}(\Cal{A})$ for all $u \in \Cal{U}$. Thus, we conclude that $\Cal{U} \subseteq \Cal{Z}(\Cal{A})$. \Par{T4.10}{(ii)}: Suppose $\Cal{D}_d(\Cal{U}) \subseteq C_{\Cal{A}}(\Cal{U})$, that is, $\Cal{D}_d(u) \in C_{\Cal{A}}(\Cal{U})$ for all $u \in \Cal{U}$. Replacing $u$ by $ru$ for any $r \in \Cal{Q}$, we have $$ d(r)u + r \Cal{D}_d(u) \in C_{\Cal{A}}(\Cal{U})\; \Rightarrow \; d(r)u \in C_{\Cal{A}}(\Cal{U}) $$ (since $C_{\Cal{A}}(\Cal{U})$ is a subalgebra of $\Cal{A}$). Therefore, $u \in C_{\Cal{A}}(\Cal{U})$ for all $u \in \Cal{U}$, which shows that $\Cal{U} \subseteq C_{\Cal{A}}(\Cal{U})$. Conversely, let $\Cal{U} \subseteq C_{\Cal{A}}(\Cal{U})$, that is, $[u,v] = 0$ for all $u, v \in \Cal{U}$. Now, by using \Par*{Theorem~4.5}, we obtain $[\Cal{D}_d(u),u] = 0$ for all $u \in \Cal{U}$. Hence $ \Cal{D}_d(\Cal{U}) \subseteq C_{\Cal{A}}(\Cal{U})$. \proclaim{Lemma 4.11} In an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$, a square-closed Lie ideal $\Cal{U}$ of $\Cal{A}$ is commutative if $$ u \circ v = 0 \quad \text{for all } u,v \in \Cal{U}. $$ \endproclaim \demo{Proof} Suppose $\Cal{U}$ be a square-closed Lie ideal of $\Cal{A}$, that means, $$ u^2 \in \Cal{U} \quad\text{and}\quad uv + vu \in \Cal{U} \quad \text{for all }u,v \in \Cal{U}. $$ If $u \circ v = 0$ for all $u,v \in \Cal{U}$, then we can write $[u,v] + 2vu = 0$. Replacing $v$ by $[v, \Cal{D}_d(v)]$, we find $$ [u, [v, \Cal{D}_d(v)]] + 2[v, \Cal{D}_d(v)] u = 0. \tag4.15 $$ Again, by putting $\Cal{D}_d(v) = v \Cal{D}_d(v)$, we obtain $$ \align [u, [v, v\Cal{D}_d(v)]] + 2[v, v\Cal{D}_d(v)] u = 0 & \Rightarrow [u, v [v, \Cal{D}_d(v)]] + 2v [v, \Cal{D}_d(v)] u = 0 \\ & \Rightarrow [u,v][v, \Cal{D}_d(v)] + v ([u,[v, \Cal{D}_d(v)]] + 2 [v, \Cal{D}_d(v)] u) = 0. \endalign $$ By \Tag(4.15), it follows that $[u,v][v, \Cal{D}_d(v)] = 0$. If $[u,v]=0$, then the result holds. On the other hand, if $[v,\Cal{D}_d(v)]=0$ for all $v\in \Cal{U}$, then \Par*{Theorem 4.5} concludes that $\Cal{U}$ is commutative. Now, we restrict the $d$-derivation $\Cal{D}_d$ to a Lie ideal $\ \Cal{U}$ and hypothesize it in different forms, such as homomorphism, antihomomorphism, and Jordan homomorphism to evaluate differential identities in connection with the commutativity of $\Cal{U}$. \proclaim{Theorem 4.12} Let $\Cal{D}_d$ be a derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$. If $\Cal{D}_d$ acts as a homomorphism {\rm(}or antihomomorphism{\rm)} on a square-closed Lie ideal $\Cal{U}$ of $\Cal{A}$, then $ \Cal{U} \subseteq \Cal{Z}(\Cal{A})$, that means, $\Cal{U}$ is central on $\Cal{A}$. \endproclaim \demo{Proof} Suppose $\Cal{D}_d$ acts as a homomorphism on $\Cal{U}$. In a square-closed Lie ideal $\Cal{U}$ of $\Cal{A}$, we have $ u^2 \in \Cal{U}$ for all $u \in \Cal{U}$. Therefore, $$ uv + vu = (u+v)^2 - u^2 - v^2 \in \Cal{U} \quad \text{for all } u, v \in \Cal{U}. $$ So, $2uv \in \Cal{U}$ for all $u, v \in \Cal{U}$. Let $[u,a], [v,b] \in \Cal{U}$ for all $u,v \in \Cal{U}$ and $a,b \in \Cal{A}$. By the homomorphism property of $\Cal{D}_d$, we have $$ \Cal{D}_d(2[u,a][v,b]) = 2 \Cal{D}_d([u,a]) \Cal{D}_d([v,b]). \tag4.16 $$ Again, using the derivation property, we obtain $$ \Cal{D}_d(2[u,a][v,b]) = 2 ( \Cal{D}_d([u,a])[v,b] + [u,a]\Cal{D}_d([v,b]) ). \tag4.17 $$ Combining \Tag(4.16) and \Tag(4.17), we get $$ 2 \Cal{D}_d([u,a]) \Cal{D}_d([v,b]) - 2 ( \Cal{D}_d([u,a])[v,b] + [u,a]\Cal{D}_d([v,b]) ) = 0. \tag4.18 $$ Replacing $u$ by $r_1 u$, where $r_1 \in \Cal{Q}$, we have $$ 2 \Cal{D}_d([r_1 u, a]) \Cal{D}_d([v,b]) - 2 ( \Cal{D}_d([r_1 u,a])[v,b] + [r_1 u,a]\Cal{D}_d([v,b]) ) = 0. $$ This implies that $$ 2 d(r_1)[u,a]\Cal{D}_d([v,b]) + 2 r_1 \Cal{D}_d([u,a]) \Cal{D}_d([v,b]) - 2 d(r_1)[u,a][v,b] - 2 r_1 \Cal{D}_d([u,a])[v,b] - 2 r_1 [u,a]\Cal{D}_d([v,b]) = 0. $$ Using \Tag(4.18) and $\operatorname{char}(\Cal{R}) \neq 2$, we get $$ [u,a]\Cal{D}_d([v,b]) - [u,a][v,b] = 0. \tag4.19 $$ Finally, setting $v$ by $r_2 v$ for any $r_2 \in \Cal{Q}$, we obtain $$ [u,a]\Cal{D}_d([r_2 v,b]) - [u,a][r_2 v,b] = 0 \Rightarrow d(r_2)[u,a][v,b] + r_2 ( [u,a]\Cal{D}_d([v,b]) - [u,a][v,b] ) = 0. $$ Using \Tag(4.19), this reduce to $$ d(r_2)[u,a][v,b] = 0,\text{ which implies } [u,a][v,b] = 0 \text{ for all }u,v \in \Cal{U} \text{ and }a,b \in \Cal{A}. $$ This suggest that $\Cal{U} \subseteq \Cal{Z}(\Cal{A})$. For the other case of $\Cal{D}_d$, proof %\? follows easily. \proclaim{Theorem 4.13} If $\Cal{D}_{d}$ acts as a Jordan homomorphism on a square closed Lie ideal $\Cal{U}$ of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$, then $\Cal{U}$ is commutative. \endproclaim \demo{Proof} Since $\Cal{D}_{d}$ acts as a Jordan homomorphism on the square closed Lie ideal $\Cal{U}$, we have $$ \Cal{D}_{d}(u \circ v) = \Cal{D}_{d}(u) \circ \Cal{D}_{d}(v)\quad\text{for all }u,v\in \Cal{U}, $$ that means, $$ \Cal{D}_{d}(uv + vu) = \Cal{D}_{d}(u)\Cal{D}_{d}(v) + \Cal{D}_{d}(v)\Cal{D}_{d}(u). $$ On the other hand, $$ \Cal{D}_{d}(uv+vu) = \Cal{D}_{d}(u)v + u\Cal{D}_{d}(v) + \Cal{D}_{d}(v)u + v\Cal{D}_{d}(u). $$ It implies that $$ \Cal{D}_{d}(u)v + u\Cal{D}_{d}(v) + \Cal{D}_{d}(v)u + v\Cal{D}_{d}(u) - \Cal{D}_{d}(u)\Cal{D}_{d}(v) - \Cal{D}_{d}(v)\Cal{D}_{d}(u) = 0. $$ Replacing $u$ by $r_{1}u$, where $r_{1}\in \Cal{Q}$ and then rearranging, we have $$ uv - u\Cal{D}_{d}(v) + vu - \Cal{D}_{d}(v)u = 0. $$ Substituting $v$ by $r_{2}v$ for any $r_{2}\in \Cal{Q}$, it reduces to $$uv + vu = 0,\quad\text{that means, } u \circ v = 0\quad\text{for all }u,v\in \Cal{U}. $$ Thus, \Par*{Lemma~4.11} implicates that $\Cal{U}$ is commutative. Now we allow the concepts of co-commuting and co-centralizing for two distinct $d$-derivations, say $d_{1}$-derivation $\Cal{D}_{d_{1}}$ and $d_{2}$-derivation $\Cal{D}_{d_{2}}$ on $\Cal{A}$ and then derive the following outcomes. \proclaim{Theorem 4.14}\Label{T4.14} In an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$, let $\Cal{U}$ be a square closed Lie ideal of $\Cal{A}$ and $\Cal{D}_{d_{1}}$, $\Cal{D}_{d_{2}}$ be respectively $d_{1}$-derivation and $d_{2}$-derivation on $\Cal{A}$. Then the following holds: \Item (i) If $\Cal{D}_{d_{1}}$ and $\Cal{D}_{d_{2}}$ are co-commuting on $\Cal{U}$, then $d_{1}$ and $d_{2}$ must be co-commuting on $\Cal{R}$. \Item (ii) If $\Cal{D}_{d_{1}}$ and $\Cal{D}_{d_{2}}$ are co-centralizing on $\Cal{U}$, but $d_{1}$ and $d_{2}$ are not co-commuting on $\Cal{R}$, then $\Cal{U}$ is commutative. \endproclaim \demo{Proof} \Par{T4.14}{(i)}: Since $\Cal{D}_{d_{1}}$ and $\Cal{D}_{d_{2}}$ are co-commuting on $\Cal{U}$, $$ \Cal{D}_{d_{1}}(u)u - u\,\Cal{D}_{d_{2}}(u) = 0 \quad \text{for all }u\in \Cal{U}. \tag4.20 $$ Replacing $u$ by $ru$ for any $r\in \Cal{R}$, we obtain $$ \Cal{D}_{d_1}(ru)ru-ru\Cal{D}_{d_2}(ru)=0 \Rightarrow d_{1}(r)ru^{2} - r d_{2}(r)u^{2} + r^{2}(\Cal{D}_{d_{1}}(u)u - u\Cal{D}_{d_{2}}(u)) = 0. $$ Using \Tag(4.20), this gives $$ (d_{1}(r)r - r d_{2}(r))u^{2} = 0. $$ Since $u^{2} \neq 0$, it follows that $$ d_{1}(r)r - r d_{2}(r) = 0 \quad\text{for all } r\in \Cal{R}. $$ This reflects that $d_{1}$ and $d_{2}$ are co-commuting on $\Cal{R}$. \Par{T4.14}{(ii)}: If $\Cal{D}_{d_{1}}$ and $\Cal{D}_{d_{2}}$ are co-centralizing on $\Cal{U}$, then $$ \Cal{D}_{d_{1}}(u)u - u\,\Cal{D}_{d_{2}}(u) \in \Cal{Z}(\Cal{A})\quad \text{for all }u\in \Cal{U}. \tag4.21 $$ Replacing $u$ by $ru$ for any $r\in \Cal{R}$, $$ d_{1}(r)ru^{2} - r d_{2}(r)u^{2} + r^{2}(\Cal{D}_{d_{1}}(u)u - u\Cal{D}_{d_{2}}(u)) \in \Cal{Z}(\Cal{A}). $$ Using \Tag(4.21), this reduces to $$ (d_{1}(r)r - r d_{2}(r))u^{2} \in \Cal{Z}(\Cal{A}). $$ Therefore for all $a\in \Cal{A}$, $$ \align [(d_{1}(r)r &- r d_{2}(r))u^{2}, a] = 0 \\ & \Rightarrow (d_{1}(r)r - r d_{2}(r))[u^{2}, a] = 0 \\ & \Rightarrow (d_{1}(r)r - r d_{2}(r)) ([u,a]u + u[u,a] )= 0 \\ & \Rightarrow (d_{1}(r)r - r d_{2}(r))([u,a]\circ u) = 0. \endalign $$ So, either $d_{1}(r)r - r d_{2}(r) = 0$ or $[u,a]\circ u = 0$. But, $d_{1}$ and $d_{2}$ are not co-commuting on $\Cal{R}$. It follows that $$ [u,a]\circ u = 0\quad \text{for all }u\in \Cal{U} \text{ and }a\in \Cal{A}. $$ Hence, using \Par*{Lemma 4.11}, we implicates that $\Cal{U}$ is commutative. \proclaim{Theorem 4.15} Let $\Cal{D}_{d}$ be a $d$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ with $\operatorname{char}(\Cal{R}) \neq 2$. For a Lie ideal $\Cal{U}$ of $\Cal{A}$ such that $\ \Cal{U} \cap \Cal{Z}(\Cal{A}) = \{0\}$, $$ C_{\Cal{A}}(\Cal{D}_{d}(\Cal{U}))=\Cal{Z}(\Cal{A})=C_{\Cal{A}}(\Cal{U}). $$ \endproclaim \demo{Proof} Let $x \in C_{\Cal{A}}(\Cal{D}_{d}(\Cal{U}))$. Then $[\Cal{D}_{d}(u),x]=0\ \text{for all } u\in \Cal{U}$. Replacing $u$ by $ua-au$ for all $a\in \Cal{A}$, we obtain $$ [\Cal{D}_{d}(ua-au),x]=0. $$ Expanding and rearranging, we get $$ \align [\Cal{D}_d(u)a,&x]+[u\Cal{D}_d(a),x]-[\Cal{D}_d(a)u,x]-[a\Cal{D}_d(u),x]=0 \\ & \Rightarrow [\Cal{D}_d(u),x]a+\Cal{D}_d(u)[a,x]+u[\Cal{D}_d(a),x]+[u,x]\Cal{D}_d(a)-[\Cal{D}_d(a),x]u \\ &\qquad-\Cal{D}_d(a)[u,x]-a[\Cal{D}_d(u),x]-[a,x]\Cal{D}_d(u)=0 \\ & \Rightarrow \Cal{D}_d(u)[a,x]+u[\Cal{D}_d(a),x]+[u,x]\Cal{D}_d(a)-[\Cal{D}_d(a),x]u-\Cal{D}_d(a)[u,x]-[a,x]\Cal{D}_d(u)=0 \\ & \Rightarrow [\Cal{D}_d(u),[a,x]] +[u,[\Cal{D}_d(a),x]]+[[u,x],\Cal{D}_d(a)]=0. \endalign $$ Again, putting $u=ru$, where $r\in \Cal{Q}$, we have $$ \aligned [\Cal{D}_d(ru),&[a,x]] +[ru,[\Cal{D}_d(a),x]]+[[ru,x],\Cal{D}_d(a)]=0 \\ & \Rightarrow d(r)[u,[a,x]]+r([\Cal{D}_d(u),[a,x]] +[u,[\Cal{D}_d(a),x]]+[[u,x],\Cal{D}_d(a)])=0 \\ & \Rightarrow [u,[a,x]]=0 \quad \text{for all } u\in\Cal{U}\quad \text{(using $d(r)\neq 0)$}. \endaligned \tag4.22 $$ Hence $[x,a] \in C_{\Cal{A}}(\Cal{U})$. Now, replacing $a$ by $\Cal{D}_{d}(a)a$ in \Tag(4.22), we obtain $$ \aligned [u,&[\Cal{D}_d(a)a,x]]=0 \\ & \Rightarrow [u,[\Cal{D}_d(a),x]a]+[u,\Cal{D}_d(a)[a,x]]=0 \\ & \Rightarrow [u,[\Cal{D}_{d}(a),x]]a+[\Cal{D}_{d}(a),x][u,a]+[u,\Cal{D}_{d}(a)][a,x]+\Cal{D}_{d}(a)[u,[a,x]]=0. \endaligned \tag4.23 $$ Finally, setting $a$ by $r_1a$, where $r_1\in \Cal{Q}$, we have $$ [u,[\Cal{D}_d(r_1a),x]]r_1a+[\Cal{D}_d(r_1a),x][u,r_1a]+[u,\Cal{D}_d(r_1a)][r_1a,x]+\Cal{D}_d(r_1a)[u,[r_1a,x]]=0. $$ Expanding, $$ \align d(r_1)&r_1[u,[a,x]]a+d(r_1)r_1[a,x][u,a]+d(r_1)r_1[u,a][a,x]+d(r_1)r_1a[u,[a,x]] \\ &\qquad + {r_1}^2 ([u,[\Cal{D}_d(a),x]]a+[\Cal{D}_d(a),x][u,a]+[u,\Cal{D}_d(a)][a,x]+\Cal{D}_d(a)[u,[a,x]] )=0 \\ & \Rightarrow d(r_1)r_1([a,x][u,a]+[u,a][a,x])=0\quad \text{(using (4.23))}. \endalign $$ Since $[x,a]\in C_{\Cal{A}}(\Cal{U})$ and $\operatorname{char}(\Cal{R})\neq 2$, it reduces to $[a,x][u,a]=0$. Therefore, either $[a,x]=0$ or $[u,a]=0$. As $\Cal{U}$ is noncentral, $[u,a]\neq 0$. Hence $[a,x]=0$ for all $a\in \Cal{A}$, that means, $x\in \Cal{Z}(\Cal{A})$. Therefore $C_\Cal{A}(\Cal{D}_d(\Cal{U}))\subseteq \Cal{Z}(\Cal{A})$. On the other hand, it is obvious that $\Cal{Z}(\Cal{A})\subseteq C_\Cal{A}(\Cal{D}_d(\Cal{U}))$, which concludes $$ C_\Cal{A}(\Cal{D}_d(\Cal{U}))=\Cal{Z}(\Cal{A}). \tag4.24 $$ Again, let $x\in C_{\Cal{A}}(\Cal{U})$. Then $[x,u]=0 \text{ for all } u\in \Cal{U}$. Applying $\Cal{D}_{d}$ on both sides and expanding, we obtain $$ [\Cal{D}_{d}(x),u]+[x,\Cal{D}_{d}(u)]=0. \tag4.25 $$ Now, replacing $u$ by $ua-au$ for all $a\in \Cal{A}$, we get $$ \aligned [\Cal{D}_d(x),&ua-au]+[x,\Cal{D}_d(ua-au)]=0 \\ &\Rightarrow [\Cal{D}_d(x), ua]-[\Cal{D}_d(x),au]+[x,\Cal{D}_d(u)a]+[x,u\Cal{D}_d(a)]-[x,\Cal{D}_d(a)u]-[x,a\Cal{D}_d(u)]=0 \\ & \Rightarrow [\Cal{D}_d(x), u]a+u[\Cal{D}_d(x),a]-a[\Cal{D}_d(x),u]-[\Cal{D}_d(x),a]u+\Cal{D}_d(u)[x,a]+[x,\Cal{D}_d(u)]a+u[x,\Cal{D}_d(a)] \\ &\qquad+[x,u]\Cal{D}_d(a)-[x,\Cal{D}_d(a)]u-\Cal{D}_d(a)[x,u]-a[x,\Cal{D}_d(u)]-[x,a]\Cal{D}_d(u)=0 \\ & \Rightarrow [u,[\Cal{D}_{d}(x),a]]+[\Cal{D}_{d}(u),[x,a]]+[u,[x,\Cal{D}_{d}(a)]]=0 \qquad \text{(using (4.25))}. \endaligned \tag4.26 $$ Putting $x$ by $rx$ for any $r\in \Cal{Q}$, we have $$ \align [u,&[\Cal{D}_d(rx),a]]+[\Cal{D}_d(u),[rx,a]]+[u,[rx,\Cal{D}_d(a)]]=0 \\ & \Rightarrow d(r)[u,[x,a]]+r ([u,[\Cal{D}_d(x),a]]+[\Cal{D}_d(u),[x,a]]+[u,[x,\Cal{D}_d(a)]] )=0. \endalign $$ Using \Tag(4.26) and $d(r)\neq 0$, we obtain $$ [u,[x,a]]=0 \quad\text{for all }u\in \Cal{U}. \tag4.27 $$ Thus $[x,a]\in C_{\Cal{A}}(\Cal{U})$. Lastly, replacing $a$ by $\Cal{D}_d(a)a$ in \Tag(4.27), we conclude that $x\in \Cal{Z}(\Cal{A})$ (same as above). Therefore, $C_{\Cal{A}}(\Cal{U})=\Cal{Z}(\Cal{A})$. Using \Tag(4.24), finally we conclude that $$ C_{\Cal{A}}(\Cal{U})=\Cal{Z}(\Cal{A})=C_{\Cal{A}}(\Cal{D}_{d}(\Cal{U})). $$ In the next result, we extend the Posner's %\? second theorem [8] for Lie ideals and discuss it in $rp$-algebra $\Cal{A}$ in the following manner. \proclaim{Theorem 4.16 \rm [8]} Let $d_{1}$ and $d_{2}$ be two ring derivations of a prime ring $\Cal{R}$ with $\operatorname{char}(\Cal{R})\neq 2$ such that the product $d_{1}d_{2}$ is also a derivation of $\Cal{R}$. Then either $d_{1}=O$ or $d_{2}=O$. \endproclaim \proclaim{Theorem 4.17} Let $\Cal{D}_{d_{1}}$ and $\Cal{D}_{d_{2}}$ be respectively $d_{1}$-derivation and $d_{2}$-derivation on an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$ and $\Cal{U}$ be a Lie ideal of $\Cal{A}$ such that $\Cal{U} \cap \Cal{Z}(\Cal{A}) = \{0\}$. If $$ \Cal{D}_{d_{1}}\Cal{D}_{d_{2}}(\Cal{U})=\{ 0 \}, $$ then $\Cal{D}_{d_{2}}=O$ on $\Cal{U}$. \endproclaim \demo{Proof} Suppose $\Cal{D}_{d_1}\Cal{D}_{d_2}(\Cal{U})=\{ 0 \}$, that means, $$ \Cal{D}_{d_1}\Cal{D}_{d_2}(u)=0\quad\text{for all }u\in \Cal{U} \tag4.28 $$ Replacing $u$ by $[u,a]$ for all $a\in \Cal{A}$, we obtain $$ \aligned \Cal{D}_{d_1}&([\Cal{D}_{d_2}(u),a]+[u,\Cal{D}_{d_2}(a)])=0 \\ & \Rightarrow [\Cal{D}_{d_1}\Cal{D}_{d_2}(u),a]+[\Cal{D}_{d_2}(u),\Cal{D}_{d_1}(a)]+[\Cal{D}_{d_1}(u),\Cal{D}_{d_2}(a)]+[u,\Cal{D}_{d_1}\Cal{D}_{d_2}(a)]=0 \\ & \Rightarrow [\Cal{D}_{d_2}(u),\Cal{D}_{d_1}(a)]+[\Cal{D}_{d_1}(u),\Cal{D}_{d_2}(a)]+[u,\Cal{D}_{d_1}\Cal{D}_{d_2}(a)]=0\ \ \ \ \text{(using (4.28))}. \endaligned \tag4.29 $$ Now, we set $\Cal{D}_{d_1}(a)=\Cal{D}_{d_1}(a)u$ and $\Cal{D}_{d_2}(a)=\Cal{D}_{d_2}(a)u$ and substituting in \Tag(4.29), we obtain $$ \align [\Cal{D}_{d_2}&(u),\Cal{D}_{d_1}(a)u]+[\Cal{D}_{d_1}(u),\Cal{D}_{d_2}(a)u]+[u,\Cal{D}_{d_1}(\Cal{D}_{d_2}(a)u)]=0 \\ & \Rightarrow [\Cal{D}_{d_2}(u),\Cal{D}_{d_1}(a)]u+\Cal{D}_{d_1}(a)[\Cal{D}_{d_2}(u),u]+[\Cal{D}_{d_1}(u), \Cal{D}_{d_2}(a)]u+\Cal{D}_{d_2}(a)[\Cal{D}_{d_1}(u),u]+[u,\Cal{D}_{d_1}\Cal{D}_{d_2}(a)]u \\ &\qquad +[u,\Cal{D}_{d_2}(a)]\Cal{D}_{d_1}(u)+\Cal{D}_{d_2}(a)[u,\Cal{D}_{d_1}(u)]=0. \endalign $$ Using \Tag(4.29) and rearranging, we get $$ \Cal{D}_{d_1}(a)[\Cal{D}_{d_2}(u),u]+[u,\Cal{D}_{d_2}(a)]\Cal{D}_{d_1}(u)=0. \tag4.30 $$ Again, substitute $u$ by $ru$ for any $r\in \Cal{Q}$, $$ \align d_2(r)&r\Cal{D}_{d_1}(a)[u,u]+r^2\Cal{D}_{d_1}(a)[\Cal{D}_{d_2}(u),u]+rd_1(r)[u,\Cal{D}_{d_2}(a)]u+r^2[u,\Cal{D}_{d_2}(a)]\Cal{D}_{d_1}(u)=0 \\ & \Rightarrow r^2 (\Cal{D}_{d_1}(a)[\Cal{D}_{d_2}(u),u]+[u,\Cal{D}_{d_2}(a)]\Cal{D}_{d_1}(u) )+rd_1(r)[u,\Cal{D}_{d_2}(a)]u=0 \\ & \Rightarrow rd_1(r)[u,\Cal{D}_{d_2}(a)]u=0\quad\text{(using (4.30))} \\ & \Rightarrow d_1(r)[u,\Cal{D}_{d_2}(a)]u=0\quad\ \text{(since }r\in \Cal{Q},\ r\neq 0). \endalign $$ Therefore $[u,\Cal{D}_{d_2}(a)]u=0$ for all $u\in \Cal{U}$. It implies that $$ [u,\Cal{D}_{d_2}(a)]=0 \quad \text{for all $a\in \Cal{A}$} . \tag4.31 $$ Finally, setting $a$ by $ua$ for all $u\in \Cal{U}$, we obtain $[u,\Cal{D}_{d_2}(ua)]=0$, which gives $$ \align [u,\Cal{D}_{d_2}&(u)a+u\Cal{D}_{d_2}(a)]=0 \\ & \Rightarrow [u,\Cal{D}_{d_2}(u)]a+\Cal{D}_{d_2}(u)[u,a]+u[u,\Cal{D}_{d_2}(a)]=0 \\ & \Rightarrow \Cal{D}_{d_2}(u)[u,a]=0\quad \text{for all }a\in \Cal{A} \quad (\text{using }(4.31)). \endalign $$ But for Lie ideal $\Cal{U}$ with $\Cal{U} \cap \Cal{Z}(\Cal{A}) = \{0\}$, we finally conclude that $\Cal{D}_{d_2}(u)=0$ for all $u\in \Cal{U}$. Therefore $\Cal{D}_{d_2}=O$ on $\Cal{U}$. At the end, we conclude our research work with the Herstein's %\? theorem [9] relating to Lie ideals using $d_1$ and $d_2$- derivations in $rp$-algebra accordingly: \proclaim{Theorem 4.18} Let $\Cal{U}$ be a noncentral Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. For $d_1$-derivation $\Cal{D}_{d_1}$ and $d_2$-derivation $\Cal{D}_{d_2}$ on $\Cal{A}$, if $$ [\Cal{D}_{d_{1}}(u),\Cal{D}_{d_{2}}(v)]=0 \quad \text{for all } u,v\in \Cal{U}, $$ then either $\Cal{D}_{d_{1}}$ or $\Cal{D}_{d_{2}}$ is commuting on $\Cal{U}$. \endproclaim \demo{Proof} Suppose $[\Cal{D}_{d_{1}}(u),\Cal{D}_{d_{2}}(v)]=0$ for all $u,v\in \Cal{U}$. Replacing $u$ by $[u,a]$ for all $a\in \Cal{A}$ and then expanding, we obtain $$ \align [\Cal{D}_{d_1}&(ua),\Cal{D}_{d_2}(v)]-[\Cal{D}_{d_1}(au),\Cal{D}_{d_2}(v)]=0 \\ & \Rightarrow [\Cal{D}_{d_1}(u)a,\Cal{D}_{d_2}(v)]+[u\Cal{D}_{d_1}(a),\Cal{D}_{d_2}(v)]-[\Cal{D}_{d_1}(a)u,\Cal{D}_{d_2}(v)]-[a\Cal{D}_{d_1}(u),\Cal{D}_{d_2}(v)]=0 \\ & \Rightarrow \Cal{D}_{d_1}(u)[a,\Cal{D}_{d_2}(v)]+u[\Cal{D}_{d_1}(a),\Cal{D}_{d_2}(v)]+[u,\Cal{D}_{d_2}(v)]\Cal{D}_{d_1}(a)-[\Cal{D}_{d_1}(a),\Cal{D}_{d_2}(v)]u \\ &\qquad-\Cal{D}_{d_1}(a)[u,\Cal{D}_{d_2}(v)]-[a,\Cal{D}_{d_2}(v)]\Cal{D}_{d_1}(u)=0. \endalign $$ After simplification, $$ [\Cal{D}_{d_{1}}(u),[a,\Cal{D}_{d_{2}}(v)]]+[u,[\Cal{D}_{d_{1}}(a),\Cal{D}_{d_{2}}(v)]]+[[u,\Cal{D}_{d_{2}}(v)],\Cal{D}_{d_{1}}(a)]=0. \tag4.31 %\?уже есть (4.31), в оригинале здесь 4.28 $$ Again, setting $u$ by $ru$ for $r\in \Cal{Q}$, we get $$ [\Cal{D}_{d_1}(ru),[a,\Cal{D}_{d_2}(v)]]+[ru,[\Cal{D}_{d_1}(a),\Cal{D}_{d_2}(v)]]+[[ru,\Cal{D}_{d_2}(v)],\Cal{D}_{d_1}(a)]=0. $$ Expanding, $$ \aligned d_1(r)[u,[a,&\Cal{D}_{d_2}(v)]] +r ([\Cal{D}_{d_1}(u),[a,\Cal{D}_{d_2}(v)]]+[u,[\Cal{D}_{d_1}(a),\Cal{D}_{d_2}(v)]]+[[u,\Cal{D}_{d_2}(v)],\Cal{D}_{d_1}(a)] )=0 \\ &\Rightarrow d_1(r)[u,[a,\Cal{D}_{d_2}(v)]]=0, \text{ which implies }\ [u,[a,\Cal{D}_{d_{2}}(v)]]=0. \endaligned \tag4.32 %\? $$ Finally, substitute $a=a\Cal{D}_{d_1}(u)$, we get $$ \align [u,[a\Cal{D}_{d_1}&(u),\Cal{D}_{d_2}(v)]]=0 \\ &\Rightarrow [u,[a,\Cal{D}_{d_2}(v)]\Cal{D}_{d_1}(u)]+[u,a[\Cal{D}_{d_1}(u),\Cal{D}_{d_2}(v)]]=0 \\ &\Rightarrow [a,\Cal{D}_{d_2}(v)][u,\Cal{D}_{d_1}(u)]+[u,[a,\Cal{D}_{d_2}(v)]]\Cal{D}_{d_1}(u)=0 \\ & \Rightarrow [a,\Cal{D}_{d_{2}}(v)][u,\Cal{D}_{d_{1}}(u)]=0\qquad \text{(using (4.35))}. \endalign $$ In $rp$-algebra, it follows that, either $[a,\Cal{D}_{d_{2}}(v)]=0$ or $[u,\Cal{D}_{d_{1}}(u)]=0$. {\sc Case 1:} If $[u,\Cal{D}_{d_{1}}(u)]=0$ for all $u\in \Cal{U}$, then \Par*{Theorem 4.5} concludes that $\Cal{D}_{d_{1}}$ is commuting on $\Cal{U}$. {\sc Case 2:} Otherwise in $[a,\Cal{D}_{d_{2}}(v)]$, replacing $a$ by $va$, we have $[va,\Cal{D}_{d_2}(v)]=0$, which gives $$ v[a,\Cal{D}_{d_2}(v)]+[v,\Cal{D}_{d_2}(v)]a=0 \Rightarrow [v,\Cal{D}_{d_{2}}(v)]a=0\quad \text{ for all $a\in \Cal{A}$}. $$ Thus, it follows that $[v,\Cal{D}_{d_{2}}(v)]=0$ for all $v\in \Cal{U}$. This leads to the conclusion that $\Cal{D}_{d_2}$ is commuting on~$\Cal{U}$. \proclaim{Corollary 4.19} Let $\Cal{U}$ be a noncentral Lie ideal of an $rp$-algebra $\Cal{A}$ over a ring $\Cal{R}$. For $d_1$-derivation $\Cal{D}_{d_1}$ and $d_2$-derivation $\Cal{D}_{d_2}$ on $\Cal{A}$, if $$ [\Cal{D}_{d_{1}}(u),\Cal{D}_{d_{2}}(v)]=0 \quad \text{ for all } u,v\in \Cal{U}, $$ then $\Cal{U}$ is commutative. \endproclaim \acknowledgment The authors express their sincere gratitude to the University of North Bengal for providing the necessary facilities and support to carry out this research work. 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