Multiplicative Lie-Type Derivations on Rings

Abstract

Let ${\mathcal {R}}$ be a ring with the center ${\mathcal {Z}}({\mathcal {R}})$ containing a nontrivial idempotent. Suppose $p_n(X_1,X_2,\dots , X_n)$ is the polynomial defined by n noncommuting indeterminates $X_1, \dots , X_n$ and their multiple Lie products. In this article, under a lenient condition on ${\mathcal {R}}$, it is shown that if a mapping $L : {\mathcal {R}} \rightarrow {\mathcal {R}}$ satisfies $L(p_n(A_{1},A_{2},\dots ,A_{n}))= \sum _{k=1}^n p_n(A_1,\dots , A_{k-1}, L(A_k), A_{k+1},\dots , A_n)$, for all $A_{1},A_{2},\dots ,A_{n} \in {\mathcal {R}}$ and $n \ge 2$ be a fixed positive integer, then for all $A,B \in {\mathcal {R}}$, there exists $Z_{A,B}$ (depending on A and B) in ${\mathcal {Z}}({\mathcal {R}})$ such that $L(A+B)=L(A)+L(B)+Z_{A,B}$.

1 Introduction

Let ${\mathcal {R}}$ be a ring with the center ${\mathcal {Z}}({\mathcal {R}})$. An additive mapping $L:{\mathcal {R}}\rightarrow {\mathcal {R}}$ is said to be a derivation on ${\mathcal {R}}$ if $L(AB)=L(A)B+AL(B)$ for all $A, B\in {\mathcal {R}}$. An additive mapping $L:{\mathcal {R}}\rightarrow {\mathcal {R}}$ is said to be a Lie derivation if

$$\begin{aligned} L([A, B]) = [L(A), B] + [A, L(B)] \end{aligned}$$

for all $A, B\in {\mathcal {R}}$, where $[A,B]=AB-BA$ is the usual Lie product. Similarly, an additive mapping $L:{\mathcal {R}}\rightarrow {\mathcal {R}}$ is said to be a Lie triple derivation if

$$\begin{aligned} L([[A,B],C])=[[L(A),B],C]+[[A,L(B)],C]+[[A,B],L(C)] \end{aligned}$$

for all $A, B, C \in {\mathcal {R}}$. Lie derivation and Lie triple derivation are also known as Lie 2-derivation and Lie 3-derivation, respectively.

It can be easily seen that every derivation on ${\mathcal {R}}$ is a Lie derivation on ${\mathcal {R}}$ and every Lie derivation on ${\mathcal {R}}$ is a Lie triple derivation on ${\mathcal {R}}$. However, the converse is not true in general. Note that if the mapping $L:{\mathcal {R}}\rightarrow {\mathcal {R}}$ is not necessarily additive in the above definitions, then L is said to be multiplicative derivation, multiplicative Lie derivation, and multiplicative Lie triple derivation on ${\mathcal {R}}$, respectively.

The question that up to what extent the multiplicative structure of a ring determines its additive structure has been considered by many researchers over the past decade. In particular, various authors have investigated the condition on ${\mathcal {R}}$ under which bijective mappings between rings preserving the multiplicative structure necessarily preserve the additive structure as well. The most fundamental result in this direction is due to Martindale III [17] who proved that every bijective multiplicative mapping from a prime ring containing a nontrivial idempotent onto an arbitrary ring is necessarily additive. Later, a number of authors considered the Lie-type product and proved that, on certain associative algebras or rings, bijective mappings which preserve any of those products are automatically additive, see [1, 3,4,5,6, 8, 12, 13, 20, 21].

Given the consideration of Lie derivations and Lie triple derivations on an algebra, Fošner et al. [9] further developed them in a more general way. Suppose that $n\ge 2$ is a fixed positive integer and $X_1,X_2,\dots ,X_n \in {\mathcal {R}}$. Let us consider a sequence of polynomials in ${\mathcal {R}}$ as:

$$\begin{aligned} p_1(X_1)= & {} X_1,\\ p_2(X_1, X_2)= & {} [p_1(X_1), X_2] =[X_1, X_2],\\ p_3(X_1, X_2, X_3)= & {} [p_2(X_1, X_2), X_3]=[[X_1, X_2], X_3],\\ p_4(X_1, X_2, X_3, X_4)= & {} [p_3(X_1, X_2, X_3), X_4]=[[[X_1, X_2], X_3], X_4],\\&\vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ , \\ p_n(X_1, X_2,\dots , X_n)= & {} [p_{n-1}(X_1, X_2,\dots , X_{n-1}), X_n].\\ \end{aligned}$$

A multiplicative Lie n-derivation is a mapping $L: {\mathcal {R}} \longrightarrow {\mathcal {R}}$ satisfying the condition

$$\begin{aligned} L(p_n(A_1, A_2,\dots , A_n))=\sum _{k=1}^n p_n(A_1,\dots , A_{k-1}, L(A_k), A_{k+1},\dots , A_n) \end{aligned}$$

for all $A_1,A_2,\dots , A_n\in {\mathcal {R}}$. Lie 2-derivations, Lie 3-derivations and Lie n-derivations are collectively referred to as Lie-type derivations.

Lie-type derivations in different backgrounds are extensively studied by several authors. Yu and Zhang [21] proved that every multiplicative Lie derivation of triangular algebras is the sum of an additive derivation and a map into its center sending commutators to zero. This result is extended to the case of multiplicative Lie triple derivations by Ji et al. [11]. Jing and Lu [12] investigated multiplicative Lie derivation on prime rings. This result is extended to multiplicative Lie triple derivation on rings by Lie et al. [13]. They Proved that every multiplicative Lie triple derivation $\delta$ from ring ${\mathcal {R}}$ into itself is nearly additive and $\delta$ is in the standard form. Fošner et al. [9] described multiplicative Lie-type derivations of von Neumann algebras and proved that every multiplicative Lie n-derivation of von Neumann algebra has the standard form. Very recently, Lin [14] gave the characterization of multiplicative generalized Lie n-derivations on triangular algebras. Many researchers have made considerable interesting works to the related topics (see [2, 7, 9, 10, 15, 16, 18,19,20]).

Motivated by the afore-mentioned works, we investigate multiplicative Lie-type derivations on rings. In fact, we prove that every multiplicative Lie-type derivation $L:{\mathcal {R}} \rightarrow {\mathcal {R}}$ is nearly additive.

2 Multiplicative Lie-Type Derivation

In this section, we examine the additivity of multiplicative Lie type derivations on rings. Suppose that ${\mathcal {R}}$ is a ring with a nontrivial idempotent $P_1$. Then, by the Pierce decomposition of ${\mathcal {R}}$, we have ${\mathcal {R}}=P_1{\mathcal {R}}P_1+P_1{\mathcal {R}}P_2+P_2{\mathcal {R}}P_1+P_2{\mathcal {R}}P_2$, where $P_2=1-P_1$. Here, the abbreviated notations $P_1{\mathcal {R}}P_1, P_1{\mathcal {R}}P_2, P_2{\mathcal {R}}P_1$ and $P_2{\mathcal {R}}P_2$ stand for the set $\{P_{1}rP_{1}\, | \, r \in {\mathcal {R}}\}, \{P_{1}r-P_{1}rP_{1}) \, | \, r \in {\mathcal {R}}\}, \{rP_{1}-P_{1}rP_{1} \, | \, r \in {\mathcal {R}}\}$ and $\{r-P_{1}r-rP_{1}+P_{1}rP_{1}\, | \,r \in {\mathcal {R}}\}$, respectively. Let us denote $P_i{\mathcal {R}}P_j={\mathcal {R}}_{ij}$ for any $i,j=1,2$. Then, ${\mathcal {R}}$ can be written as ${\mathcal {R}}={\mathcal {R}}_{11}+{\mathcal {R}}_{12}+{\mathcal {R}}_{21}+{\mathcal {R}}_{22}$. Throughout this paper, $A_{ij}$ will denote an arbitrary element of ${\mathcal {R}}_{ij}$ and any element $A \in {\mathcal {R}}$ can be expressed as $A=A_{11}+A_{12}+A_{21}+A_{22}$. Recall that a ring ${\mathcal {R}}$ is prime if $A{\mathcal {R}}B=\{0\}$ implies that either $A=0$ or $B=0$, and is semiprime if $A{\mathcal {R}}A=\{0\}$ implies $A=0$. The main result of our paper states as follows.

Theorem 2.1

Let $n\ge 2$ and ${\mathcal {R}}$ be a ring containing a nontrivial idempotent $P_1$ and satisfying the following condition

(S) If $A_{11}B_{12}=B_{12}A_{22}$ for all $B_{12} \in {\mathcal {R}}_{12}$, then $A_{11}+A_{22} \in {\mathcal {Z}}({\mathcal {R}})$.

Suppose that a mapping $L : {\mathcal {R}} \rightarrow {\mathcal {R}}$ satisfies

$$\begin{aligned} L(p_n(A_{1},A_{2},\dots ,A_{n}))= \sum _{k=1}^n p_n(A_1,\dots , A_{k-1}, L(A_k), A_{k+1},\dots , A_n) \end{aligned}$$

for all $A_{1},A_{2},\dots ,A_{n} \in {\mathcal {R}}$. Then, for all $A,B \in {\mathcal {R}}$, there exists $Z_{A,B}$ (depending on A and B) in ${\mathcal {Z}}({\mathcal {R}})$ such that $L(A+B)=L(A)+L(B)+Z_{A,B}$.

In proving the upcoming lemmas, we shall use the hypothesis of Theorem 2.1 freely, without any specific mention.

Lemma 2.2

$L(0)=0$.

Proof

Indeed,

$$\begin{aligned} L(0)= & {} L(p_n(0,0,\dots , 0))\\= & {} p_n(L(0),0,\dots , 0)+p_n(0,L(0),\dots , 0)+\dots +p_n(0,0,\dots , L(0))\\= & {} 0. \end{aligned}$$

$\square$

Lemma 2.3

For any $A_{11} \in {\mathcal {R}}_{11}, B_{12} \in {\mathcal {R}}_{12}$ and $B_{21} \in {\mathcal {R}}_{21}$, there exists $Z_{A_{11},B_{12}}, Z_{A_{11},B_{21}} \in {\mathcal {Z}}({\mathcal {R}})$ such that

(i):

$L(A_{11}+B_{12})=L(A_{11})+L(B_{12})+Z_{A_{11},B_{12}}$,

(ii):

$L(A_{11}+B_{21})=L(A_{11})+L(B_{21})+Z_{A_{11},B_{21}}$.

Proof

We prove $\mathrm{(i)}$, the proof of $\mathrm{(ii)}$ follows in a similar manner. Let $M=L(A_{11}+B_{12})-L(A_{11})-L(B_{12})$. Since $p_n(A_{11},P_1,\dots , P_1)=0$ and $p_n(A_{11}+B_{12},P_1,\dots , P_1)=p_n(B_{12},P_1,\dots , P_1)$, we have

$$\begin{aligned}&L(p_n(A_{11}+B_{12},P_1,\dots , P_1))\\&\quad =p_n(L(A_{11}+B_{12}),P_1,\dots , P_1)+p_n(A_{11}+B_{12},L(P_1),\dots , P_1)\\&\qquad +\dots +p_n(A_{11}+B_{12},P_1,\dots , L(P_1)). \end{aligned}$$

On the other hand, using Lemma 2.2, we have

$$\begin{aligned}&L(p_n(A_{11}+B_{12},P_1,\dots , P_1))= L(p_n(A_{11},P_1,\dots , P_1))+L(p_n(B_{12},P_1,\dots , P_1))\\&\quad =p_n(L(A_{11}),P_1,\dots , P_1)+p_n(A_{11},L(P_1),\dots , P_1)\\&\qquad +\dots +p_n(A_{11},P_1,\dots , L(P_1))\\&\qquad +p_n(L(B_{12}),P_1,\dots , P_1)+p_n(B_{12},L(P_1),\dots , P_1)\\&\qquad +\dots +p_n(B_{12},P_1,\dots , L(P_1)). \end{aligned}$$

Comparing the above two relations, we obtain

$$\begin{aligned} p_n(M,P_1,\dots , P_1)=0. \end{aligned}$$

That is,

$$\begin{aligned} P_1MP_2=0=P_2MP_1. \end{aligned}$$

(2.1)

For any $C_{12} \in {\mathcal {R}}_{12}$, on one hand, we have

$$\begin{aligned}&L(p_n(A_{11}+B_{12},C_{12},P_1,\dots , P_1))\\&\quad =p_n(L(A_{11}+B_{12}),C_{12},P_1,\dots , P_1)+p_n(A_{11}+B_{12},L(C_{12}),P_1,\dots , P_1)\\&\qquad +\dots +p_n(A_{11}+B_{12},C_{12},P_1,\dots , L(P_1)). \end{aligned}$$

On the other hand, we have

$$\begin{aligned}&L(p_n(A_{11}+B_{12},C_{12},P_1,\dots , P_1))\\&\quad =L(p_n(A_{11},C_{12},P_1,\dots , P_1))+L(p_n(B_{12},C_{12},P_1,\dots , P_1))\\&\quad =p_n(L(A_{11}),C_{12},P_1,\dots , P_1)+p_n(A_{11},L(C_{12}),P_1,\dots , P_1)\\&\qquad +\cdots +p_n(A_{11},C_{12},P_1,\dots , L(P_1))+p_n(L(B_{12}),C_{12},P_1,\dots , P_1)\\&\qquad +p_n(B_{12},L(C_{12}),P_1,\dots , P_1)+\dots +p_n(B_{12},C_{12},P_1,\dots , L(P_1)). \end{aligned}$$

Comparing the above two relations, we obtain $p_n(M,C_{12},P_1,\dots ,P_1)=0$, i.e., $P_1MP_1C_{12}=C_{12}P_2MP_2$ for all $C_{12} \in {\mathcal {R}}_{12}$. By the condition S, we have

$$\begin{aligned} P_1MP_1+P_2MP_2 \in {\mathcal {Z}}({\mathcal {R}}). \end{aligned}$$

(2.2)

Now, (2.1) together with (2.2) implies that $L(A_{11}+B_{12})=L(A_{11})+L(B_{12})+Z_{A_{11},B_{12}}$ for some $Z_{A_{11},B_{12}} \in {\mathcal {Z}}({\mathcal {R}})$. $\square$

Lemma 2.4

For any $A_{12}, B_{12} \in {\mathcal {R}}_{12}$ and $A_{21}, B_{21} \in {\mathcal {R}}_{21}$, we have

(i):

$L(A_{12}+B_{12})=L(A_{12})+L(B_{12})$,

(ii):

$L(A_{21}+B_{21})=L(A_{21})+L(B_{21})$.

Proof

Using Lemma 2.3, we compute

$$\begin{aligned} L(A_{12}+B_{12})= & {} L(p_n(P_1-B_{12},P_1+A_{12},-P_1,\dots , -P_1))\\= & {} p_n(L(P_1-B_{12}),P_1+A_{12},-P_1,\dots , -P_1)\\&\quad +p_n(P_1-B_{12},L(P_1+A_{12}),-P_1,\dots , -P_1)\\&\quad +\cdots +p_n(P_1-B_{12},P_1+A_{12},-P_1,\dots , L(-P_1))\\= & {} p_n(L(P_1)+L(-B_{12}),P_1+A_{12},-P_1,\dots , -P_1)\\&+p_n(P_1-B_{12},L(P_1)+L(A_{12}),-P_1,\dots , -P_1)\\&+\cdots +p_n(P_1-B_{12},P_1+A_{12},-P_1,\dots , L(-P_1))\\= & {} L(p_n(P_1,P_1,-P_1,\dots , -P_1))+L(p_n(P_1,A_{12},-P_1,\dots , -P_1))\\&+L(p_n(-B_{12},P_1,-P_1,\dots , -P_1))\\&\quad +L(p_n(-B_{12},A_{12},-P_1,\dots , -P_1))\\= & {} L(A_{12})+L(B_{12}). \end{aligned}$$

Similarly, we can prove the part $\mathrm{(ii)}$. $\square$

Lemma 2.5

For any $A_{ii}, B_{ii} \in {\mathcal {R}}_{ii}$, there exists $Z_{A_{ii},B_{ii}} \in Z({\mathcal {R}})$, such that

$$\begin{aligned} L(A_{ii}+B_{ii})=L(A_{ii})+L(B_{ii})+Z_{A_{ii},B_{ii}}. \end{aligned}$$

Proof

Let us denote $M=L(A_{11}+B_{11})-L(A_{11})-L(B_{11}).$ By Lemma 2.2, we get

$$\begin{aligned} 0= & {} L(0)=L(p_n(A_{11}+B_{11},P_{1},\dots ,P_{1}))\\= & {} p_n(L(A_{11}+B_{11}),P_{1},\dots ,P_{1}),p_n(A_{11}+B_{11},L(P_{1}),\dots ,P_{1})\\&+\cdots +p_n(A_{11}+B_{11},P_{1},\dots ,L(P_{1})). \end{aligned}$$

On the other hand, we see that

$$\begin{aligned} 0= & {} L(p_n(A_{11},P_{1},\dots ,P_{1}))+L(p_n(B_{11},P_{1},\dots ,P_{1}))\\= & {} p_n(L(A_{11}),P_{1},\dots ,P_{1})+p_n(A_{11},L(P_{1}),\dots ,P_{1})\\&+\cdots +p_n(A_{11},P_{1},\dots ,L(P_{1}))\\&+p_n(L(B_{11}),P_{1},\dots ,P_{1}) +p_n(B_{11},L(P_{1}),\dots ,P_{1})\\&+\cdots +p_n(B_{11},P_{1},\dots ,L(P_{1})). \end{aligned}$$

Comparing the above two relations, we obtain $p_n(M,P_{1},\dots ,P_{1})=0$, that is

$$\begin{aligned}&P_{1}MP_{2}=0=P_{2}MP_{1}~~~~\text{ i.e. }~~~~M_{12}=0=M_{21}. \end{aligned}$$

(2.3)

For any $C_{12} \in {\mathcal {R}}_{12}$, using Lemma 2.4, we have

$$\begin{aligned} L((A_{11}+B_{11})C_{12})= & {} L(A_{11}C_{12})+L(B_{11}C_{12})\\= & {} L(p_n(A_{11},C_{12},-P_{1},\dots ,-P_{1}))\\&+L(p_n(B_{11},C_{12},-P_{1},\dots ,-P_{1}))\\= & {} p_n(L(A_{11}),C_{12},-P_{1},\dots ,-P_{1})\\&+p_n(A_{11},L(C_{12}),-P_{1},\dots ,-P_{1})\\&+\cdots +p_n(A_{11},C_{12},-P_{1},\dots ,L(-P_{1}))\\&+p_n(L(B_{11}),C_{12},-P_{1},\dots ,-P_{1})\\&+p_n(B_{11},L(C_{12}),-P_{1},\dots ,-P_{1})\\&+\cdots +p_n(B_{11},C_{12},-P_{1},\dots ,L(-P_{1})). \end{aligned}$$

On the other hand, we have

$$\begin{aligned} L((A_{11}+B_{11})C_{12})= & {} L(p_n(A_{11}+B_{11},C_{12},-P_{1},\dots ,-P_{1}))\\= & {} p_n(L(A_{11}+B_{11}),C_{12},-P_{1},\dots ,-P_{1})\\&+p_n(A_{11}+B_{11},L(C_{12}),-P_{1},\dots ,-P_{1})\\&+\cdots +p_n(A_{11}+B_{11},C_{12},-P_{1},\dots ,L(-P_{1})). \end{aligned}$$

On comparing the above two relations, we arrive at

$$\begin{aligned}&p_n(M,C_{12},-P_{1},\dots ,-P_{1})=0.~~~~\text{ In } \text{ other } \text{ words }\\&\quad P_1MP_1C_{12}=C_{12}P_2MP_2~~~\text{ for } \text{ all }~~~C_{12} \in {\mathcal {R}}_{12}. \end{aligned}$$

In view of the condition (S), we have

$$\begin{aligned} P_1MP_1+P_2MP_2 \in {\mathcal {Z}}({\mathcal {R}}). \end{aligned}$$

(2.4)

Further combining (2.3) and (2.4), we find that $L(A_{11}+B_{11})=L(A_{11})+L(B_{11})+Z_{A_{11},B_{11}}$ for some $Z_{A_{11},B_{11}} \in {\mathcal {Z}}({\mathcal {R}})$. Similarly, we can show the case for $i=2$. $\square$

Lemma 2.6

For any $A_{12}\in {\mathcal {R}}_{12}$ and $B_{21}\in {\mathcal {R}}_{21}$, we have

$$\begin{aligned} L(A_{12}+B_{21})=L(A_{12})+L(B_{21}). \end{aligned}$$

Proof

Using Lemma 2.3, we have

$$\begin{aligned}&L((-1)^{n+1}A_{12}+B_{21})\\&\quad =L(p_n(P_{1}+A_{12},P_{1}-B_{21},P_{1},\dots ,P_1))\\&\quad =p_n(L(P_{1}+A_{12}),P_{1}-B_{21},P_{1},\dots ,P_{1})\\&\qquad +p_n(P_{1}+A_{12},L(P_{1}-B_{21}),P_{1},\dots ,P_{1})\\&\qquad +\cdots +p_n(P_{1}+A_{12},P_{1}-B_{21},P_{1},\dots ,L(P_{1}))\\&\quad =p_n(L(P_{1})+L(A_{12}),P_{1}-B_{21},P_{1},\dots ,P_{1})\\&+p_n(P_{1}+A_{12},L(P_{1})-L(B_{21}),P_{1},\dots ,P_{1})\\&\qquad +\cdots +p_n(P_{1}+A_{12},P_{1}-B_{21},P_{1},\dots ,L(P_{1}))\\&\quad =L(p_n(P_{1},P_{1},P_{1},\dots ,P_1))+L(p_n(P_{1},-B_{21},P_{1},\dots ,P_1))\\&\qquad +L(p_n(A_{12},P_{1},P_{1},\ dots,P_1))+L(p_n(A_{12},-B_{21},P_{1},\dots ,P_1))\\&\quad =L((-1)^{n+1}A_{12})+L(B_{21}). \end{aligned}$$

$\square$

Lemma 2.7

For any $A_{11}\in {\mathcal {R}}_{11}$, $B_{12}\in {\mathcal {R}}_{12}$, $C_{21}\in {\mathcal {R}}_{21}$, and $F_{22}\in {\mathcal {R}}_{22}$, we have

$$\begin{aligned}&L(A_{11}+B_{12}+C_{21}+F_{22})=L(A_{11})+L(B_{12})\\&+L(C_{21})+L(F_{22})+Z_{A_{11},B_{12},C_{21},F_{22}}. \end{aligned}$$

Proof

Let $M=L(A_{11}+B_{12}+C_{21}+F_{22})-L(A_{11})-L(B_{12})-L(C_{21})-L(F_{22})$. By Lemmas 2.2 and 2.6, we find that

$$\begin{aligned}&p_n(L(A_{11}+B_{12}+C_{21}+F_{22}),P_1,\dots ,P_1)+p_n(A_{11}+B_{12}+C_{21}\\&\qquad +F_{22},L(P_1),\dots ,P_1)+\cdots +p_n(A_{11}+B_{12}+C_{21}+F_{22},P_1,\dots ,L(P_1))\\&\quad =L(p_n(A_{11}+B_{12}+C_{21}+F_{22},P_1,\dots ,P_1))\\&\quad =L((-1)^{n+1}B_{12}+C_{21})=L((-1)^{n+1}B_{12})+L(C_{21})\\&\quad =L(p_n(A_{11},P_{1},P_{1},\dots ,P_1))+L(p_n(B_{12},P_{1},P_{1},\dots ,P_1))\\&\qquad +L(p_n(C_{21},P_{1},P_{1},\dots ,P_1))+L(p_n(F_{22},P_{1},P_{1},\dots ,P_1))\\&\quad =p_n(L(A_{11}),P_{1},\dots ,P_{1})+p_n(A_{11},L(P_{1}),\dots ,P_{1})\\&\qquad +\cdots +p_n(A_{11},P_{1},\dots ,L(P_{1}))\\&\qquad +p_n(L(B_{12}),P_{1},\dots ,P_{1})+p_n(B_{12},L(P_{1}),\dots ,P_{1})\\&\qquad +\cdots +p_n(B_{12},P_{1},\dots ,L(P_{1}))\\&\qquad +p_n(L(C_{21}),P_{1},\dots ,P_{1})+p_n(C_{21},L(P_{1}),\dots ,P_{1})\\&\qquad +\cdots +p_n(C_{21},P_{1},\dots ,L(P_{1}))\\&\qquad +p_n(L(F_{22}),P_{1},\dots ,P_{1})+p_n(F_{22},L(P_{1}),\dots ,P_{1})\\&\qquad +\cdots +p_n(F_{22},P_{1},\dots ,L(P_{1}))\\&\quad =p_n(L(A_{11})+L(B_{12})+L(C_{21})+L(F_{22}),P_{1},\dots ,P_{1})\\&\qquad +p_n(A_{11}+B_{12}+C_{21}+F_{22},L(P_{1}),\dots ,P_{1})\\&\qquad +\cdots +p_n(A_{11}+B_{12}+C_{21}+F_{22},P_{1},\dots ,L(P_{1})). \end{aligned}$$

Hence, it follows that $p_n(M,P_1,\dots ,P_1)=0$, that is

$$\begin{aligned} P_1MP_2=0=P_2MP_1. \end{aligned}$$

(2.5)

For any $X_{12} \in {\mathcal {R}}_{12}$, by Lemmas 2.2 and 2.4, we obtain

$$\begin{aligned}&p_n(L(A_{11}+B_{12}+C_{21}+F_{22}),X_{12},P_1,\dots ,P_1)+p_n(A_{11}+B_{12}+C_{21}\\&\qquad +F_{22},L(X_{12}),P_1,\dots ,P_1)+\cdots +p_n(A_{11}\\&\qquad +B_{12}+C_{21}+F_{22},X_{12},P_1,\dots ,L(P_1))\\&\quad =L(p_n(A_{11}+B_{12}+C_{21}+F_{22},X_{12},P_1,\dots ,P_1))\\&\quad =L((-1)^{n+1}(X_{12}F_{22}-A_{11}X_{12}))\\&\quad =L((-1)^{n+1}(X_{12}F_{22}))+L((-1)^{n+1}(-A_{11}X_{12}))\\&\quad =L(p_n(A_{11},X_{12},P_1,\dots ,P_1))+L(p_n(B_{12},X_{12},P_1,\dots ,P_1))\\&\qquad +L(p_n(C_{21},X_{12},P_1,\dots ,P_1))+L(p_n(F_{22},X_{12},P_1,\dots ,P_1))\\&\quad =p_n(L(A_{11}),X_{12},P_{1},\dots ,P_{1})+p_n(A_{11},L(X_{12}),P_{1},\dots ,P_{1})\\&\qquad +\cdots +p_n(A_{11},X_{12},P_{1},\dots ,L(P_{1}))\\&\qquad +p_n(L(B_{12}),X_{12},P_{1},\dots ,P_{1})+p_n(B_{12},L(X_{12}),P_{1},\dots ,P_{1})+\dots \\&\qquad +p_n(B_{12},X_{12},P_{1},\dots ,L(P_{1}))+p_n(L(C_{21}),X_{12},P_{1},\dots ,P_{1})\\&\qquad +p_n(C_{21},L(X_{12}),P_{1},\dots ,P_{1})+\cdots +p_n(C_{21},X_{12},P_{1},\dots ,L(P_{1}))\\&\qquad +p_n(L(F_{22}),X_{12},P_{1},\dots ,P_{1})+p_n(F_{22},L(X_{12}),P_{1},\dots ,P_{1})\\&\qquad +\cdots +p_n(F_{22},X_{12},P_{1},\dots ,L(P_{1}))\\&\quad =p_n(L(A_{11})+L(B_{12})+L(C_{21})+L(F_{22}),X_{12},P_{1},\dots ,P_{1})\\&\qquad +p_n(A_{11}+B_{12}+C_{21}+F_{22},L(X_{12}),P_{1},\dots ,P_{1})\\&\qquad +\cdots +p_n(A_{11}+B_{12}+C_{21}+F_{22},X_{12},P_{1},\dots ,L(P_{1})). \end{aligned}$$

This leads to $L(p_n(M,X_{12},P_1,\dots ,P_1))$, which implies $P_1MP_1X_{12}=X_{12}P_2MP_2$ for all $X_{12} \in {\mathcal {R}}_{12}$. Hence, by the condition (S), we have

$$\begin{aligned} P_1MP_1+P_2MP_2 \in {\mathcal {Z}}({\mathcal {R}}). \end{aligned}$$

(2.6)

Now, (2.5) together with (2.6) implies that $M \in Z({\mathcal {R}})$. Therefore, $L(A_{11}+B_{12}+C_{21}+F_{22})=L(A_{11})+L(B_{12})+L(C_{21})+L(F_{22})+Z_{A_{11},B_{12},C_{21},F_{22}}$ for some $Z_{A_{11},B_{12},C_{21},F_{22}} \in {\mathcal {Z}}({\mathcal {R}})$. $\square$

Now, we are ready to prove our main result.

Proof of Theorem 2.1

For any $a, b \in {\mathcal {R}}$, we write $a=A_{11}+A_{12}+A_{21}+A_{22}$ and $b=B_{11}+B_{12}+B_{21}+B_{22}$. Applying Lemmas 2.4, 2.5, and 2.7, we arrive at

$$\begin{aligned} L(a+b)= & {} L(A_{11}+A_{12}+A_{21}+A_{22}+B_{11}+B_{12}+B_{21}+B_{22})\\= & {} L((A_{11}+B_{11})+(A_{12}+B_{12})+(A_{21}+B_{21})+(A_{22}+B_{22}))\\= & {} L(A_{11}+B_{11})+L(A_{12}+B_{12})+L(A_{21}+B_{21})+L(A_{22}+B_{22})+Z_{1}\\= & {} L(A_{11})+L(B_{11})+Z_{2}+L(A_{12})+L(B_{12})+L(A_{21})+L(B_{21})\\&\quad +L(A_{22})+L(B_{22})+Z_{3}+Z_{1}\\= & {} L(A_{11}+A_{12}+A_{21}+A_{22})-Z_{4}+L(B_{11}+B_{12}+B_{21}+B_{22})-Z_{5}\\&\quad +Z_{1}+Z_{2}+Z_{3}\\= & {} L(A)+L(B)+Z_{A,B}, \end{aligned}$$

where $Z_{A,B}=Z_{1}+Z_{2}+Z_{3}-Z_{4}-Z_{5}$.

$\square$

Now, we apply Theorem 2.1 to prime rings, triangular algebras, and nest algebras. We begin with the following lemma.

Lemma 2.8

Let ${\mathcal {R}}$ be a prime ring containing a nontrivial idempotent. If $A_{11}B_{12}=B_{12}A_{22}$ for all $B_{12}\in {\mathcal {R}}_{12}$, then $A_{11}+A_{22}\in {\mathcal {Z}}({\mathcal {R}})$.

Proof

For any $B_{11}\in {\mathcal {R}}_{11}$ and $B_{12}\in {\mathcal {R}}_{12}$, we get $A_{11}B_{11}B_{12}=B_{11}B_{12}A_{22}=B_{11}A_{11}B_{12}$ for all $B_{12}\in {\mathcal {R}}_{12}$. As ${\mathcal {R}}$ is prime, we have $A_{11}B_{11}=B_{11}A_{11}.$ For any $B_{12}\in {\mathcal {R}}_{12}$ and $B_{22}\in {\mathcal {R}}_{22}$, we get $B_{12}B_{22}A_{22}=A_{11}B_{12}B_{22}=B_{12}A_{22}B_{22}$ for all $B_{12}\in {\mathcal {R}}_{12}$. Therefore, by the primeness of ${\mathcal {R}}$ that $B_{22}A_{22}=A_{22}B_{22}$. For any $B_{12}\in {\mathcal {R}}_{12}$ and $B_{21}\in {\mathcal {R}}_{21}$, we get $A_{22}B_{21}B_{12}=B_{21}B_{12}A_{22}=B_{21}A_{11}B_{12}$ for all $B_{12}\in {\mathcal {R}}_{12}$. It follows that $A_{22}B_{21}=B_{21}A_{11}$. For any $B\in {\mathcal {R}}$, we have

$$\begin{aligned} (A_{11}+A_{22})B= & {} (A_{11}+A_{22})(B_{11}+B_{12}+B_{21}+B_{22})\\= & {} A_{11}B_{11}+A_{11}B_{12}+A_{22}B_{21}+A_{22}B_{22}\\= & {} B_{11}A_{11}+B_{12}A_{22}+B_{21}A_{11}+B_{22}A_{22}\\= & {} (B_{11}+B_{12}+B_{21}+B_{22})(A_{11}+A_{22})\\= & {} B(A_{11}+A_{22}). \end{aligned}$$

Hence, we find that $A_{11}+A_{22}\in {\mathcal {Z}}({\mathcal {R}})$. $\square$

It follows from Lemma 2.8 that every prime ring with a nontrivial idempotent satisfies the condition of Theorem 2.1. Hence, we have the following corollary.

Corollary 2.9

Let $n\ge 2$ and ${\mathcal {R}}$ be a prime ring containing a nontrivial idempotent. Suppose that a mapping $L : {\mathcal {R}} \rightarrow {\mathcal {R}}$ (not necessarily additive) satisfying

$$\begin{aligned} L(p_n(A_{1},A_{2},\dots ,A_{n}))= \sum _{k=1}^n p_n(A_1,\dots , A_{k-1}, L(A_k), A_{k+1},\dots , A_n) \end{aligned}$$

for all $A_{1},A_{2},\dots ,A_{n} \in {\mathcal {R}}$. Then, for all $A,B \in {\mathcal {R}}$, there exists $Z_{A,B}$ (depending on A and B) in ${\mathcal {Z}}({\mathcal {R}})$ such that $L(A+B)=L(A)+L(B)+Z_{A,B}$.